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DESCRIPTIVE GEOMETRY 





Coal Age v Electric Railway Journal 
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Metallurgical & Chemical Engineering 
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DESCRIPTIVE GEOMETRY 



BY 

WILLIAM L. AMES 

FORMERLY PROFESSOR OF MACHINE DESIGN, 
ROSE POLYTECHNIC INSTITUTE 



AND 



CARL WISCHMEYER 

ASSOCIATE PROFESSOR OF MACHINE DESIGN, 
ROSE POLYTECHNIC INSTITUTE 



Fifth Edition 



MCGRAW-HILL BOOK COMPANY, Inc. 
239 WEST 39TH STREET. NEW YORK 



LONDON: HILL PUBLISHING CO., Ltd. 
6 & 8 BOUVERIE ST., E. C. 

1918 



f\s~ 
''fit 



Copyright, 1918, by the 
McGraw-Hill Book Company, Inc. 



X H B /m A P ^ K PRESS YORK PA 

Jy)eU5U3783 

o I 



PREFACE 

The book " Notes on Descriptive Geometry " by William 
L. Ames, was first published in 1893, and was probably the 
earliest book on the subject which used the third quadrant. 
Even today, when the third quadrant is used exclusively 
in the drafting offices of this country, the subject of 
Descriptive Geometry is to a considerable extent taught 
as first quadrant projection. 

The present book is in reality little more than an en- 
largement of Professor Ames' book, the principal changes 
being in the increased number of exercises. Additional 
illustrations have been put in and a few chapters added. 
In this work of revision and enlargement the author desires 
to express his indebtedness to Professor John B. Peddle 
of the Rose Polytechnic Institute. 

C. W. 

Terre Haute, Ind. 
Sept. 1, 1918. 



CONTENTS 



Page 

Preface v 

Chapter 

I. Introduction 1 

II. Representation of Points 3 

III. Representation of Lines 6 

IV. Representation of Planes 10 

V. Representation of Simple Solids 12 

VI. Assuming New Planes' of Projection 13 

VII. Change of Position by Rotation 17 

VIII. Lines with Reference to H and V 25 

IX. Planes with Reference to H and V , 31 

X. Development of Surfaces 36 

XL Line Contained in Plane 40 

XII. Intersections 50 

XIII. Perpendiculars 56 

XIV. Angles 61 

XV. Distances 64 

XVI. Tangent Planes 67 

XVII. Shades and Shadows ' 72 

XVIII. Sections 83 

XIX. Intersections of Surfaces 88 

XX. Helicoidal Surfaces 94 

XXI. Hyperboloid of Revolution of One Nappe 97 

XXII. Practical Applications 99 

XXIII. Impossible and Indeterminate Exercises Ill 



DESCRIPTIVE GEOMETRY 



I. Introduction. 

Descriptive Geometry treats of methods of representing 
on plane surfaces, magnitudes of three dimensions in such 
a way that their forms and positions may be completely 
determined; and conversely, of methods of determining the 
forms and positions of magnitudes thus represented. 

If the intersection with a plane of the rays from all points 
of an object to the observer is taken as the representation 
of the object on the plane, the production of such a repre- 
sentation requires that three things be known, namely the 
position of the object, of the plane of representation and of 
the observer. 

In Fig. 1, let p represent a point in space and o the posi- 
tion of the observer. A line drawn 
from one to the other will intersect 
the vertical plane at p v , and we 
say that p v is the projection of the 
point p upon this plane. In like 
manner suppose the position of 
the observer to be shifted to o' 
and again connected to p by a 
straight line. The intersection of the horizontal plane and 
this line gives us the point p h , which is called the projec- 
tion of p upon the horizontal plane. 

Thus knowing the position of the point p in space with 
reference to the planes of representation, and the position 
of the observer, or the direction of a line from the point to 

1 




Fig. 1. 



DESCRIPTIVE GEOMETRY 



„—- -^ 2nd. Quad 



the observer, we are able to find the projections of the 
point. 

Or, if we reverse the process and have given the point 
p v as the projection of the point p upon the vertical plane, 
when the observer is at o, we know that p must lie some- 
where upon the line op v extended, though its exact position 
on this line is indeterminate. Also if p h is the projection 
of p upon the horizontal plane, when o' is the observer's 
position, we know that p lies somewhere upon o'p h extended. 
Now having p v and p h with the directions op v and o f p } \ the 
location of the point p in space is fully determined with 
reference to the planes of representation. 

This method is that used in Descriptive Geometry, but 

simplified by the following 



modifications : The planes of 
representation, which are also 
used as planes of reference as 
to position, are taken at right 
angles to each other and are 
considered as unlimited in 
extent. The points of ob- 
servation are taken to be at 
an infinite distance away, so that the rays become lines per- 
pendicular to the planes of projection and the representa- 
tions are called orthographic projections of the object in 
space. 

One of the planes is taken vertical and called the vertical 
plane of projection or V; the other is taken horizontal and 
called the horizontal plane of projection or H. The two 
planes intersect in a line called the ground line and desig- 
nated by X. The intersection of the two planes forms 
four dihedral angles called quadrants. The space above 
H and in front of V is called the first quadrant, above H 
and behind V the second, below H and behind V the third, 
and below H and in front of V the fourth (Fig. 2) . 




Fig. 2. 



REPRESENTATION OF POINTS 3 

For the convenience of the draftsman the planes are 
considered as rotated about X, so that the first and third 
quadrants open to 180° and the second and fourth quad- 
rants close to 0°. 

II. Representation of Points. 

In considering the representation of a point as modified 
by the above conditions, X is taken as a horizontal line, 
and indicates the line of intersection of H and V. It also 
represents the axis about which H and V are rotated into 
one plane which is represented by the plane of the paper. 
All the representations or projections which, when the 
planes are at right angles, would be in V above H or in H 
back of V, will appear in the drawing 
above X. All projections which would 
be in V below H or in H in front of V, will 
appear in the drawing below X. Hence it 
will be seen that if a point is in the first 
quadrant its projection on the V plane, 
called its V projection or elevation, will be 
above X, and its projection on the H plane, 
called its H projection or plan, will be below X. If the 
point lies in the second quadrant, both projections will be 
above X; if in the third quadrant, the H projection will be 
above X and the V projection below; if in the fourth 
quadrant, both projections will be below X. 

From Fig. 1 it is seen that since p v p and p h p are respec- 
tively perpendicular to V and H, that the projection of all 
points of the line p h p on V will be the line sp v perpendicular 
to the line of intersection of H and V. Also the projection 
of the line p v p on H will be the line sp h also perpendicular 
to the line of intersection of H and V. Hence when H and 
V are revolved into the same plane, p v and p h will lie in the 
same perpendicular to X, as in Fig. 3. The distance from 
p h to X measures the distance from p to the V plane, and the 




.<7" 

/ '; 


6 V 


<" 








\C 





4 DESCRIPTIVE GEOMETRY 

distance from p v to X measures the distance from p to the 
H plane. 

Figure 4 shows the projections or representations of four 
points, one lying in each quadrant. From this it will be 
noticed that any two points lying in the same perpendicular 
to X may be taken as the projections of a point in space. 

Notation. — Points in space will be designated by the 
small letters, as a, b, c. The V projections will be desig- 
nated by the same letters with the exponent v, as a v , b v , c v , 
and the H projections by the same 
letters with the exponent h, as a h , 
b h , c h . Successive positions of the 
same points will be designated by 
subscripts, as a h a 2 , a s . 

Problem 1. — Having the direction 
and distance of a point in space from 
H and V, to draw its projections. 
Draw any line perpendicular to X and set off from X, 
above if the point is above H, or below if the point is below 
H, the distance of the point from H; this will be the V 
projection of the point. On the same perpendicular set 
off from X, above if the point is in back of V, or below if 
the point is in front of V, the distance of the point from V; 
this will be the H projection of the point. 

Problem 2. — Having one projection of a point in space 
and the direction and distance of the point from that plane of 
projection, to draw the other projection. 

A perpendicular to X through the given projection will 
contain the other projection, which will then be located as 
in Problem 1. 

Problem 3. — Having the projections of a point in space, 
to determine its position with reference to H and V. 

The distance of the H projection from X will show the 
distance of the point from V, and if the H projection is 
above X the point is back of V; if below X, in front of V. 



REPRESENTATION OF POINTS 5 

In like manner the distance of the V projection from X will 
show the distance of the point from H. If the V projection 
is above X the point is above H; if below, the point is below 
H. 

Solution of Exercises. — All exercises are to be drawn full 
scale unless otherwise noted. The exercises are designed to 
fit on standard letter size paper, 8% in. X 11 in. The H 
and V projections of a point should always be connected by 
a projecting line perpendicular to X, preferably a dotted 
line as shown in the figures. 

The universal use of the third quadrant in practical 
work makes it desirable to use this quadrant in the solution 



i 

^ | 

x-f4 






& 



TT 



n 



b» 









a* 



T? fi 



>B 



c 



ti 



d» 



lO* 



Fig. 5. 



^ 



Fig. 6. 



^i 



of exercises. In a few of the exercises the data specified 
definitely locate the object in one of the other quadrants, 
but wherever any choice is left, the third quadrant should 
be used. 

Exercises 



1. Show the projections of the following points: 

1 in. behind V, 1}4 i n - below H. 

2 in. behind V, 1 in. above H. 

3 in. in front of V, 1 in. above H. 
1 in. in front of V, 1 in. below H. 
in V, 1 in. below H. 
in V, 2 in. above H. 
1 in. behind V, in H. 
1 in. in front of V, in H. 
in V, in H. 



a, 
b, 

d, 

/, 



6 DESCRIPTIVE GEOMETRY 

2. State where each of the points in Fig. 5 is located, giving dis- 
tances and directions from H and V. 

3. Show the projections of four points, one in each quadrant, each 
1 in. from H and 2 in. from V. 

4. State in which quadrant each of the points shown in Fig. 6 is 
located, and whether the point is nearer V or H. 

5. The plan of a point is 1 in. above X, and the point is 1J4 m - 
below H. Show its projections. 

6. A point 1 in. above X is the plan of three points, a, b and c. 
a is 1 in. above H, b is in H and c is 1% m - below H. Show the 
projections of the points. 

7. Three points, a, b and c, lie in a plane perpendicular to X. a 

and b have their elevations in the 
same point, and b and c have their 
plans in the same point, b is 2 in. 
from H and V, a is J£ in. from V and 
c is 1 in. from H. Show the projec- 
tions of the points. 

8. The elevations of two points 
coincide at a point 1% in. below X. 
The points are equidistant from H and 
V, one in front of V and the other 
behind V. Show the projections of 
the points. 

9. A point in the third quadrant which had its plan J^ in. from X 
and its elevation 2 in. from X has moved 2 in. to the right and % in. 
down. Show the projections of its former and present positions. 

10. A point in the first quadrant is 1 in. from H and 2 in. from V. 
It moves }4 m - further from H, J^ in. nearer V and 1 in. to the right. 
Show the projections of its original and final positions. 

11. The point a in Fig. 7 moves about a vertical line through c in 
such a way that the plan of its path- is a circle. For each 30° that 
a h moves about c h , a moves }£ in. away from H. Show the projec- 
tions of the path of the point for one complete revolution. 

III. Representation of Lines. 

The projecting lines of all the points of a right line in 
space form its projecting plane, the intersection of which 
with the plane of projection determines the projection of 
the line in space. The projection of a line in space is there- 
fore determined by joining the corresponding projections 




REPRESENTATION OF LINES 



of any two points of the line. Hence, since any two points 
in the same perpendicular to X may be taken as the projec- 
tions of a point in space, it will be seen that any two lines 
may be taken as the projections of a line in space, provided 
that they can be cut by two perpendiculars to X. 

Notation. — Lines in space w T ill be designated by capital 
letters, as A, B, C. The V projections of the lines will be 
designated by the same letters with the exponent v, as A v , 
B v , C*, and the H projections by the same letters with the 
exponent h, as A h , B h , C h . A line determined by the points 
a and b will be called 
the line ab. Also the 
point determined by 
the intersection of 
the two lines C and 
D will be called the 
point CD. 

The points of in- 
tersection of a line 
with the H and V 
planes of projection 
are called its H and 
V traces, and are in- 
dicated by h and v. 

Problem 4. — Having the projections of a line, to find its traces. 

The H trace of a line is a point in the H plane, therefore 
its V projection must lie in X. Since the H trace is a 
point of the line, its V projection must lie in the V projec- 
tion of the line. If, therefore, the V projection of the line 
is extended till it cuts X, this point will be the V projection 
of the H trace. A perpendicular to X at this point will 
cross the H projection of the line at the H trace. Simi- 
larly, to find the V trace extend the H projection till it 
cuts X, then a perpendicular to X at this point will cross 
the V projection of the line at the V trace (Fig. 8). 




Fig. 8. 



8 DESCRIPTIVE GEOMETRY 

Note. — A line is said to cross a given quadrant when 
the portion between the traces lies in that quadrant. 

Since the projections of points determine the projec- 
tions of lines containing them, exercises in the projection 
of lines must depend in general on Problems 1 and 2 to 
determine two points of the line, which is then itself 
determined. 

A line parallel to H has all its points at the same distance 
from H. Therefore the V projection of the line must have 
all its points the same distance from X, or in other words 
must be parallel to X. Similarly a line parallel to V has its 
H projection parallel to X. 

With reference to H and V a line may have six general 
positions : 

Line 

A, parallel to H parallel to V 

B, parallel to H perpendicular to V 

C, parallel to H inclined to V 

D, perpendicular to H parallel to V 

E, inclined to H parallel to V 

F, inclined to H inclined to V 

H projection V projection 

A h , parallel to X A v , parallel to X 

B h , perpendicular to X B v , a point 
C h , inclined to X C v , parallel to X 

D h , a point D v , perpendicular to X 

E h , parallel to X E v , inclined to X 

F h , inclined to X F v , inclined to X 

Note an exception to F when the line is in a plane per- 
pendicular to X. The projections then coincide in a line 
perpendicular to X and the line in space is indeterminate. 

Exercises 

12. Represent a line in each of the general positions indicated above. 

13. Show the projections and traces of four lines, one crossing each 



REPRESENTATION OF LINES 9 

quadrant, the H and V projections of each line meeting X in points 
2 in. apart. 

14. Show the projections of a line 2 in. long lying in the fourth 
quadrant and perpendicular to H. One end of the line lies in H at a 
point I in. from V. 

15. Show the projections of a line parallel to X, 1 in. from H and 
2 in. from V. 

16. A line crossing the second quadrant has its H and V traces 
each 2 in. from X. Its projections meet X at points 4 in. apart. 
Show the projections of the line. 

17. Lines A and B meet at a point 1 in. from H and 2 in. from V. 
A is parallel to H and inclined at 45° to V. B is parallel to V and 
inclined at 30° to H. Show the projections and traces of the lines. 

18. Show the projections and traces of a line passing through X 
and inclined to H and V. 

19. Show the projections and traces of a line inclined at 30° to V, 
parallel to H and 1 in. from it. 

20. The H trace of a line is 2 in. below X, the V trace 3 in. to the 
right and 1 in. below X. Show the projections of the line. 

21. Show the projections and traces of the line containing the 
points a and b of Exercise 1, the projecting lines of the points being 

1 in. apart. 

22. Show the projections and traces of the line containing the 
points c and d of Exercise 1, the projecting lines of the points being 

2 in. apart. 

23. Show the projections and traces of the line containing the 
points b and c of Exercise 1, the projecting lines of the points being 
2 in. apart. 

24. Show the projections of the triangle having as vertices the 
points c, d and e of Exercise 1, the projecting lines of c and d coincid- 
ing and 1 in. from the projecting line of e. 

25. Show the projections of the triangle having as vertices the 
points a, b and c of Exercise 2, the projecting lines of b and c being 
respectively 1 in. and 23^ in. from that of a. 

26. Show the projections of all the triangles that may be formed 
having as vertices the points a, b, c and d of Exercise 1, the distances of 
the projecting lines of a to those of 6, c and d being respectively 1 in., 
2 in. and 3% in. 

27. The point AB is 1 in. from H and 1% in. from V; the H projec- 
tion of this point and the H traces of the lines A and B form an equi- 
lateral triangle of 1J^ in. side, of which one side is inclined at 45° to 
X. Find the V traces of A and B. 



10 



DESCRIPTIVE GEOMETRY 



28. The point a is at the H trace and the point b at the V trace of 
the line ab. The elevation of the line is inclined at 30° and the plan 
at 45° to X. The distance a v b v is 2 in. Show the projections of the 
line. 

IV. Representation of Planes. 

Planes are generally represented by their lines of inter- 
section with H and V, called H and V traces, forming 




Fig. 9. 

two lines intersecting in X and extending indefinitely on 
either side of it (Fig. 9) . 

For the sake of clearness it is customary to represent 
only the part of the plane included in one quadrant. The 
plane shown in Fig. 9 can be represented in four different 
ways, according as the part of the plane assumed is taken 

in the first, second, third or 
fourth quadrant (Fig. 10). 

In the following, unless 
otherwise stated, planes will 
be taken in the third quad- 
rant. The H trace is there- 
fore represented above and 
the.V trace below X. 

Notatio N. — Planes, ex- 
cepting the planes of projection, will be designated by the 
numerals, as 1 , 2, 3; the V trace by the same numeral with 
the exponent v, as l v , 2 V , 3 V ; the H trace by the same 




Fig. 10. 



REPRESENTATION OF PLANES 11 

numeral with the exponent h, as l h , 2 h , 3 h . The plane 
determined by the points a, b and c will be called the 
plane abc, and its traces abc h and abc v . The plane 
determined by two intersecting lines A and B will be 
called the plane AB, and its traces AB h and AB V . The 
plane determined by the line A and the point b will be called 
the plane Ab and the traces Ab h and Ab v . The line deter- 
mined by the intersection of the planes 1 and 2 will be called 
the line 12, and the projections 12 h and 12 v . The point 
determined by the intersection of the planes 1, 2 and 3 will 
be called the point 123, and the projections 123 h and 123 v . 
The point determined by the intersection of the plane 1 and 
the line A will be called the point 1A and the projections 
TA h and TA V . 

With reference to H and V a plane may have eight 
general positions: 

Plane 

1, perpendicular to H perpendicular to V 

2, inclined to H perpendicular to V 

3, perpendicular to H inclined to V 

4, inclined to H inclined to V 

5, perpendicular to H parallel to V 

6, parallel to H perpendicular to V 

7, parallel to, but not intersecting X 

8, passing through X 

H trace V trace 

l h , perpendicular to X l v , perpendicular to X 

2 h , perpendicular to X 2 V , inclined to X 

3 h , inclined to X 3 V , perpendicular to X 

4 h ] inclined to X If, inclined to X 

5 h , parallel to X 5 V , at infinity 
6 h , at infinity t 6 V , parallel to X 

7 h , parallel to X 7 V , parallel to X 

8 h , in X 8 V , in X 



12 DESCRIPTIVE GEOMETRY 

Exercises 

29. Represent a plane in each of the general positions indicated 
above. 

30. Show the traces of a plane which is perpendicular to V and 
inclined at 30° to H. 

31. Inclined at 45° to V and perpendicular to H. 

32. Parallel to V and 1 in. behind it, 

33. Parallel to H and Y 2 in. below it. 

34. Parallel to H and 2 in. above it. 

35. Parallel to V and % in. in front of it. 

36. Represent the plane of Exercise 30 in the first quadrant. 

37. Show the traces of a plane parallel to X and meeting H and V 
at an angle of 45°. 

V. Representation of Simple Solids. 

Solids are represented by showing the form and relative 
position of the surfaces bounding them. These surfaces are 
in turn known when we know their determining lines. 
Hence solids bounded by plane surfaces are represented by 
the lines of intersection of these surfaces. Solids bounded 
by curved surfaces are generally represented by their 
apparent boundary lines. The solids may be placed at 
any convenient distances from the planes of projection. 
Hidden lines are represented by short dashed lines. 

Exercises 

38. Draw the plan and elevation of a rectangular prism having 
the base parallel to H and a side face parallel to V, the base being 1 in. 

X 3 in. and the altitude 2 in. 

39. Draw the projections of a cube of 2 in. edge with top and 
bottom faces parallel to H and a diagonal parallel to V. 

40. Show the projections of a triangular prism with its axis vertical 
and a side face making an angle of 30° with V. Its base is equilateral 
of 2 in. side and its altitude 3 in. 

41. Draw the projections of a cylinder 2 in. long and 2 in. in diam- 
eter, axis perpendicular to V. 

42. Show the plan and elevation of a cone of revolution having 
for its base a 3 in. circle in H, while its vertex is 2 in. below H. 

43. Draw the projections of a pyramid, the base being a square of 
2 in. side situated parallel to V and 3 in. from it. The vertex is in V. 



ASSUMING NEW PLANES OF PROJECTION 



13 



5" A 3" 

8 ^y-<-" 



8 



Fig. 11. 



44. Draw the projections of a 2 in. sphere whose center is J^ in. 
from H and 2 in. from V. 

45. Draw the projections of a cone of revolution whose base is 
2 in. in diameter and altitude 3 in. Its axis lies in V and 
its base in H. 

46. A sphere of 2 in. diameter has a vertical hole 1 in. in diameter 
bored through its center. Show its plan and elevation. 

47. A torus, or anchor ring, is generated by revolving a 1 in. 
circle about a vertical axis distant 1 in. from 
the center of the circle. Show H and V pro- 
jections. 

48. Figure 1 1 shows the plan of a stick whose 
right section is 1 inch square. Show the ele- 
vation. 

49. A cylinder 2 in. in diameter and 3 in. long has its axis parallel 
to X. Show its projections. Note that the projections do not 
fully define the shape of the cylinder. 

VI. Assuming New Planes of Projection. 

As has been noted in the representation of lines, when the 
line is in a plane perpendicular to X it is indeterminate in 
direction or position, and may be straight or curved. 
Hence it is evident that in the representation of objects 
bounded by plane surfaces, when any of these planes comes 

into a position per- 
pendicular to X, the 
outline of the surface 
cannot be determined 
from the plan and ele- 
vation. Therefore it 
is frequently neces- 
sary to show more 
than two views of a 
given object. 

As an example, the 
cylinder of Exercise 
49 is represented by two equal rectangles, of which the 
lines parallel to X will represent the sides of the cylinder 




Fig. 12. 



14 DESCRIPTIVE GEOMETRY 

and the lines perpendicular to X the ends of the cylinder. 
But these two views would represent as well a prism of 
square section, Fig. 12, so that to determine the form fully 
an end view is necessary. The new plane of projection is 
usually taken perpendicular to both H and V, and is then 
called the Profile plane and is denoted by P. The line of 
intersection of P and V is called Y, and the line of inter- 
section of P and H is called Z. P projections of points 
2\ and lines, and P traces of planes 

V'~~~^^ will be indicated by the exponent 

p, as a p , B p and 3 P . 

In some cases it is more con- 
venient to take the third plane of 
projection perpendicular to H but 
inclined to V, as for example when 
it is desired to get a true end view 
of a solid whose end face is inclined 
FlG - 13 - to V. The general method of find- 

ing the third projection is the same in either case. 

Problem 5. — Given the H and V projections of a point 
in space j to show its projection on a new vertical plane. 

To bring the new vertical plane into the plane of the 
paper it is customar}' to revolve it about its V trace or Y. 
This revolution is made in such a direction that Z moves 
clockwise into coincidence with X. The P projection of the 
point will fall the same distance below or above X as the 
V projection. To find the distance of the P projection 
to the right or left of the Y axis, draw a perpendicular 
from a h to Z (Fig. 13). The required distance will evidently 
be the distance between the foot of the perpendicular and 
the point s, and this distance can be conveniently laid off 
by drawing the arc about s as center, as shown in the 
figure. 

If the new plane of projection is perpendicular to both H 
and V, Y and Z form one straight line perpendicular to X, 



ASSUMING NEW PLANES OF PROJECTION 



15 



and the perpendicular from ah to Z becomes a horizontal 
line. This is illustrated in Fig. 14, which shows the 
projections of a triangular prism. 

Exercises 

50. Show H, V and P projections of the c}dinder of Exercise 49. 

51. Show the P projections of the points a, b and c of Exercise 1. 
62. Show the P 

projection of the line 
ab of Exercise 21. 

53. Show the P 
projection of the 
triangle cde of Exer- 
cise 24. 

54. Points a and b 
lie in a plane per 
pendicular to X. a 
is \}/2 in. from H 



ix- 




Fig. 14. 



b is 1 in. from H and 
1% in. from V. Show 
projectionsand traces 
of the line ab. 

55. Point c lies 1 in. from V, 2 in. from H and 3 in. from P. Point 
d lies 2 in. from V, 1 in. from H and 1 in. from P. Find H, V and P 
traces of the line cd- 

56. The H trace of a line is 2 in. above X and the V trace 1 in. 
below X, both traces being in the same perpendicular to X. Show 

H, V and P projections of the line. 

57. Plane 2 has its H trace inclined at 30° 
and its V trace at 45° to X. Find its P 
trace. 

58. Find the projections of a point of 
line A which is twice as far from H as it is 
from V, Fig. 15. 

59. Show the traces of a plane parallel 
to X and inclined at 30° to H. 

60. Draw the plan and elevation of a 
Fig. 15. holt 1J4 in. diameter, 3 in. long under the 

head, the head being hexagonal, \% in. 
short diameter and 1 in. thick. Take axis parallel to X. 

61. Show the projections of a square frame 4 in. outside, the sides 




16 



DESCRIPTIVE GEOMETRY 



of the frame being 1 in. square. Take a 4 in. X 4 in. face parallel 
to V. 

62. Show the projections of the frame of the preceding exercise 
when the long dimension of the plan is inclined at 30° to X. 

63. Show the projections of a hexagonal prism 2 in. long and 2 in. 
long diameter, with the axis parallel to H and inclined at 30° to V. 

64. Show top, side and end views of a bolt having a hemispherical 
head of 2 in. diameter. The shank for 1J^ in. under the head is 
1 in. square, the remaining 2 in. cylindrical and 1 in. in diameter. 
Take axis parallel to X. 

65. Project a circle of 2 in. diameter lying in a plane parallel to 
V, on a plane perpendicular to H and inclined at 60° to V. 

66. Project a cube of 1J^ in. edge on a plane perpendicular to one 





-lev, 



Fig. 16. 



Fig. 17. 



of its diagonals. Assume the position of the cube so that one of its 
diagonals is parallel to H. 

67. A cone has for its elevation an equilateral triangle of 2 in. side. 
The vertex of the cone is in H and the plane of the base makes an 
angle of 45° with H. Show plan, front and end elevations. 

68. A sphere of 2Y 2 in. diameter is pierced by a vertical hole 1}$ 
in. square. Show top and front view. 

69. In Fig. 16 is shown the H projection of a 3 in. sphere pierced 
by a vertical rectangular hole. Show V and P projections. 

70. A hexagonal prism of 3 in. length, 2 in. long diameter, has a 
cylindrical hole of % in. diameter half its length. Show plan and 
elevations with axis parallel to X. 

71. Show the projections of the box of Fig. 17 with the cover raised 
at an angle of 45°. 



CHANGE OF POSITION BY ROTATION 



17 




Fig. 18. 



72. A rectangular block lJ-£ in. X 2 in. X 3 in. has two of its 

3 in. edges in planes 2>^ in. apart and parallel to H. Show plan and 
elevation of the block. 

73. A set of four equal steps having each a rise and tread of 8 in. is 

4 ft. long. Show elevation when 
the long dimension of the plan 
is inclined at 30° to X. Scale, 
% in. = 1 ft. in. 

VII. Change of Position 
by Rotation. 

The views obtained by 
assuming new planes of 
projection can also be ob- 
tained by changing the 
position of the object with 

reference to H and V by rotation about some axis. 
When a point is revolved about an axis, it describes the 
circumference of a circle, the plane of the circle is perpendicu- 
lar to the axis, its center is in the axis and its radius is the 
shortest distance from the point to the axis. If the axis 
is oblique to the planes of projection the path of the point 
will be projected as ellipses. For ease and simplicity 
of construction the axis should be taken perpendicular to one 
of the planes of projection. The result obtained by rotating 
about an oblique axis can be obtained by two rotations 
about perpendicular axes, one perpendicular to V, the other 
toH. 

If the axis is taken perpendicular to H, then the H 
projection of the path of any point will be a circular arc 
with center at the H projection of the axis, and the V pro- 
jection of the path a straight line parallel to X and through 
the V projection of the point (Fig. 18). Correspondingly, 
if the axis is taken perpendicular to V, the V projection 
of the path of any point is an arc of a circle and the H 
projection a straight line parallel to X. 



18 



DESCRIPTIVE GEOMETRY 



Figure 19 shows the rotation of a line B about an axis 
A j the axis being perpendicular to V. This problem would 
be greatly simplified by taking the axis of rotation inter- 
secting the line B, in which case the point of intersection 
would be a stationary point. 

Figure 20 shows a solution of Exercise 66 by rotating 
the cube so that a diagonal is perpendicular to V, in which 
case the new V projection is the required projection. 

There are certain cases in which the limitations given 





Fig. 19. 



Fig. 20. 



above as to the positions of the axes are not necessary. 
(a) The revolution of a point into either plane of projec- 
tion about any line. in that plane as axis. (6) The revolu- 
tion of a line into either plane of projection about an axis 
lying in that plane and passing through the corresponding 
trace of the line, (c) The revolution of a plane into either 
plane of projection about its intersection with that plane 
as axis. 

Problem 6. — To revolve a given point into one of the planes 
of projection about an axis in that plane, the axis being inclined 
to X and containing the projection of the point. ■ 

Given point p and the axis in H, to revolve the point 



CHANGE OF POSITION BY ROTATION 



19 




Fig. 21. 



Fig. 22. 



into H (Fig. 21). Since the plane of revolution of the point 
will be perpendicular to the axis of revolution, a line drawn 
through p h perpendicular to the axis will contain the 
revolved position of the 
point. The radius of 
the circle described by 
the point is the short- 
est distance from the 
point to the axis and 
is measured by the dis- 
tance from p v to X. 
This distance set off 
from p h along the per- 
pendicular will locate the revolved position of p. 

The position of a point after revolution into one of the 
planes of projection will be designated by enclosing the 
letter in parentheses, as (p). 

Problem 7. — To revolve a given point into one of the planes 

of projection about an axis in 
that plane, the axis being in- 
clined to X and not containing 
the projection of the point. 

Given point p and the axis in 
V, to revolve the point into V 
(Fig. 22). In this case, as be- 
fore, the revolved position of 
the point will lie in a perpen- 
dicular to the axis of revolu- 
tion and passing through p v . 
The radius of the circle is 
the shortest distance from the 
point to the axis, and is found as the hypotenuse of a 
right triangle having the distances from p h to X and from 
p v to the axis as sides. This distance being found by con- 
struction is set off from the axis, giving (p) as shown. 




Fig. 23. 



20 



DESCRIPTIVE GEOMETRY 



Problem 8. — To revolve a given line into one of the planes 
of projection about an axis in that plane and passing through 
the corresponding trace of the line. 

Given line A and the axis in H passing through the H 
trace, to revolve the line into H (Fig. 23). It is evident 
that the H trace of the line, being the point of intersection 
of the line and the axis of revolution, will not change its 
position during the revolution. If any other point of the 
line is assumed, such as a, and its revolved position is 
found as in Problem 7, the revolved position of the line 




Fig. 24. 



Fig. 25. 



will be the line joining the revolved position of the point 
with the H trace of the line. 

This problem is simplified by letting the axis of revolution 
coincide with the H projection of the line (Fig. 24). The 
revolution of the assumed point will then be accomplished 
by the method of Problem 6. 

Note. — If the axis of revolution does not pass through 
the H trace of the line, that is, if the line and axis are not 
in the same plane, the problem is impossible. 

Problem 9. — To revolve a given plane into one of the planes 
of projection about its trace as axis. 



CHANGE OF POSITION BY ROTATION 



21 




Fig. 26. 



Given plane 2, to revolve it into H about its H trace 

(Fig. 25). This result is effected by taking any point of 

the V trace and revolving it about the H trace into H. The 

position of (p) may be 

found by the right triangle 

construction of Problem 7, 

but a simpler method is to 

draw an arc with center at 

s and with the distance sp v 

as radius. The point in 

which the arc cuts the 

perpendicular to 2 h from 

p h is (p), since s(p) and sp v 

must be of equal length. 

Problem 10. — To find the true distance between two given 

points. 
First Method. — Revolve the line containing the points 

into or parallel to one of the planes of projection, whence 

the projection on that plane will show the true dis- 
tance. 

Second Method. — The true distance can be found as the 

hypotenuse of a right triangle, 
one side of which is the 
length of one projection of 
the distance, the other side 
the difference in distances of 
the two points from the cor- 
responding plane of projec- 
tion. In Fig. 26, let it be 
required to find the distance 
between b and c. Construct 
a right triangle with the length 
of 6V as one side and the 

difference in distances from H, or the length x as the other 

side, then the hypotenuse is the required true distance. 




Fig. 27 



22 



DESCRIPTIVE GEOMETRY 



Problem 11. — Given of a line one projection and its true 
length, to find the other projection. 

Given A h and the true length of A (Fig. 27). Either 
method of Problem 10 can be applied. 

First Method. — Assume the V projection of one point on 
the line, as b v . Suppose the line to be revolved parallel 
to V about an axis through b, giving the new H projection as 
A± h . Then the V projection of the line will have one end at 



x- 




Fig. 28. Fig. 29. 

b v , the other end in the perpendicular dropped from Ci h , 
and will have a length equal to the given true length. 
Counter revolve the line to its original position, giving A v . 
Second Method. — Construct a right triangle with the 
length of the given H projection as one side and the true 
length as hypotenuse. The other side will be the difference 
in distances from H of the points b and c. 

Exercises 

74. Revolve p through 60° about A as axis (Fig. 28;. 

75. A regular tetrahedron of 3 in. edge has its base in a plane 
parallel to H and 3 in. from it, one side of the base being parallel to V. 
Show its projections. 

76. Revolve the tetrahedron shown in Fig. 29 through 105° about 
A as axis. 

77. Revolve B into a position parallel to V (Fig. 30). 

78. Revolve C into a position parallel to X (Fig. 31). 



CHANGE OF POSITION BY ROTATION 



23 



79. Revolve D into a position perpendicular to V (Fig. 32). 

80. A square of 2 in. side stands in a plane perpendicular to X, 
with its sides parallel to H and V. Revolve about one of the vertical 
sides through 60°, then about an axis perpendicular to V and contain- 
ing a corner of the square, through 120°. 

81. A regular hexagon of 1)4 m - side lies in a plane perpendicular 



x- 




x- 





x- 





Fig. 30. 



Fig. 31. 



Fig. 32. 



to X, with two sides horizontal. Revolve the hexagon through 45° 
about a vertical axis passing through a corner of the hexagon. 

82. Point a is in H, 2 in. behind V; point b is 1 in. to the right, 
1 in. below H and 1 in. behind V; point c is 1 in. to the right of &, 
\Yi in. below H and 2 in. behind V. Show projections and true form 
of the triangle abc. 

83. A 3 in. circle lies in a plane parallel to V and 2 in. from it. The 
circle is revolved about the vertical diameter 

as axis until its V projection is an ellipse 
with minor axis IK m - long. Show pro- 
jections of the circle before and after rota- 
tion. Find at least twelve points of the 
revolved position. 

84. The H trace of a line is 2 in, above X, 
the V trace 13^ in. below X and the line is 
6 in. long between traces. Show its pro- 
jections. 

85. Triangle def has fd = 3 in., fe — 4 in. 
and de = 5 in. d and e lie in X and / is 1 in. 
below H. Show projections of the triangle. 

86. The H trace of line A is 2 in. from X. 
X and is 3 in. long. The true length of A is 4 in. Find the V projec- 
tion and V trace of A. 

87. Show the projections of a point on line A at a true distance of 
W2 in. from the H trace (Fig. 33). 




Fig. 33. 



A h is inclined at 30° to 



24 



DESCRIPTIVE GEOMETRY 



88. The P trace of a line is 1 in. from H and 2 in. from V. The 
V trace is J£ in. from H and 1}^ m - from P. Find the true distance 
between the H and P traces. 

89. A line 2 in. long between its H and P traces has its H trace 

1 in. from V and l}i in. from P. Its P trace 
is 1 in. from H. Show its projections. 

90. A rod 2% in. long is suspended hori- 
zontally by vertical threads 3 in. long 
attached to its ends. Show how far the 
rod will be raised by turning it through 90°. 

91. A rod 2 in. long stands perpendicular 
to H at a point 2 in. from V. A second rod 

2 in. long stands perpendicular to V at a point 1 J^ in. from H. The 
distance from the foot of one rod to the foot of the other is 4 in. 
What is the distance between the other ends of the rods? 




Fig. 34. 





Fig. 36. 



Fig. 37. 



92. The plan of a triangle abc is equilateral of 2 in. side, ab = 2% 
in., be = 3 in. What is the length of ac? 

93. Of the triangle def, de = 1 in., ef = 1*4 m - an d df = 2 in. 

4 h 



x- 





X- 



T 



4 V 



Fig. 38. 



Fig. 39. 



d and / are in X and e is equidistant between H and V. Show plan 
and elevation of the triangle. 

94. Revolve E into H about E h as axis (Fig. 34). 

95. Revolve F into V about the axis shown in V (Fig. 35). 



CHANGE OF POSITION BY ROTATION 



25 



96. The line C, whose H projection is given, has been revolved into 
H about the axis, and (p) is a point on its revolved position. Find 
the V projection of C (Fig. 36). 

97. Revolve the plane 2 into H (Fig. 37). 

98. Revolve the plane 3 into V (Fig. 38). 




99. Revolve the plane 4 into H (Fig. 39). 

100. Revolve the plane 5 into H (Fig. 40). 

101. An isosceles triangle of 2 in. base and 3 in. sides lies in H with 
the base parallel to X. Revolve the triangle through 60° about one 
of the 3 in. sides. 



x- 





V of 


' 




r^4* 


^V 




Fig. 41. 






Fig. 42. 

102. Revolve line A into H about A h as axis (Fig. 41). 

103. Point a, when revolved into H about an axis in H, moves to 
(a). Find the axis (Fig. 42). 

VIII. Lines with Reference to H and V. 

A line in space has the following distinctive data with 
reference to H and V : H projection, V projection, _0/>he 



26 



DESCRIPTIVE GEOMETRY 



angle of inclination with H, and <f>, the angle of inclination 
with V. Any two of these are enough to determine the 
other two. The H and V traces of the line are important 
points, but are readily derived from the projections, or the 
projections from the traces, and in connection with other 
data have simply the effect of any other given points of 
the line. 

Problem 12. — Given the projections of a line, to find 6 and <f> 
(Fig. 43). 
First Method. — Since the angle between a line and a 

plane is the angle between the 
line and its projection on that 
plane, the angle 6 is the angle 
between the line and its H 
projection. To find its value, 
revolve the line into H about 
its H projection, whence the 
angle between the revolved 
position of the line and its 
Similarly to 
find </>, revolve about the V projection into V. 

Second Method. — The angle between the V projection of 
the line and X is a projection of the angle 6. To find its 
true value revolve the line about an axis perpendicular to 
H until the H projection is parallel to X. The angle 
between the new V projection and X is the required angle. 
Similarly to find </>, revolve the line about an axis per- 
pendicular to V until the V projection is parallel to X. 
Figure 43 illustrates both methods of finding the angle 6. 
Problem 13. — Given of a line one projection and the 
angle which the line makes with the corresponding plane of 
projection, to find the other projection. 
Given A h and 6 (Fig. 44). 

First Method. — From any point of A h draw a line (A) 
making the required angle with it. This line will represent 




Fig. 43. 

original H projection is the required angle 



LINES WITH REFERENCE TO H AND V 



27 



the line in space revolved into H about A h as axis, and the 
point chosen on A h will represent the H trace of the line. 




Fig. 44. 



Fig. 45. 



Make the counter revolution, which will give the required 
V projection. 

Second Method. — Swing A h parallel to X, draw the V 
projection making the given 
angle with X, and make the 
counter revolution. 

Problem 14. — Given of a 
line one ' projection and the /\" 
angle which the line makes 
with the other plane of projec- 
tion, to find the other projection. 

Given B v and 6 (Fig. 45). 
Consider the line revolved into 
a position parallel to V, whence 
Bi v will make the angle with 
X and Bi h will be parallel to 
X. Make the counter revo- 
lution, which will give the required H projection. The 
counter revolution may be made in either direction, giving 
two possible positions. 

Problem 15. — To find the projections of a line making 
the angle 6 -with H and the angle <j> with V. 




Fig. 46. 



28 



DESCRIPTIVE GEOMETRY 



Note.— The limits of the value of + cf> are 0° and 90°. 

First Method. — Suppose b in Fig. 46 to be one of the points 

of the required line, 
then all the lines pass- 
ing through b and 
making an angle 
with V will lie on the 
surface of a cone hav- 
ing b as vertex and 
the circle d v e v e i v d 1 v as 
a base, the line b h a h 
representing the slant 
height of the cone and 
inclined at the angle 
4> to X. Construct 
the right triangle 
b h a h c with the angle 

Fig. 47. ■ ° 

at b h equal to the 
given angle 6. Then since b h a h is the true length of every 
element of the cone, b h c will represent the length of the H 




bi 



X- 



<-. 



«_ 



T 



SCV, 

± 



projection of an element making 
the required angle with H. 
With this length as radius and 
b h as center draw the arc cutting 
X at the points d h and e h . The 
corresponding V projections will 
be d v , di v , e v and ei v , giving four 
possible positions of the line. 

This problem may be solved 
equally well by drawing a 6 cone 
with its vertex at b and base in 
H, and using the angle </> in the 
construction of the right triangle. In some cases it may 
be more convenient to place the base of the d or </> cone 
parallel to instead of in H or V. 



CM- 



Fig. 48. 



LINES WITH REFERENCE TO H AND V 



29 



Second Method (Fig. 47) . — Construct both 8 and $ cones 
as above, then an element which lies on both cones will be 
the required line. In order to find such a line it is necessary 
that the slant heights of the two cones be made equal. In 
this case the bases of the cones will be intersecting circles 
and the line joining a point of intersection with the common 
vertex b will be a solution of the problem. In order to 
get all four possible solutions it is necessary to construct 
both nappes of one of the cones. 

Exercises 

104. Find and </> for the line ~ab, Fig. 48. 

105. A line having its H trace 1^ in. from X meets V 2 in. below 
X. Find 6 when <j> is 30°. 

106. A line has its H projection inclined at 30° to X. Its V trace 
is 2 in. below X. 6 is 30°, Show projections and traces of the line. 





Fig. 49. 



Fig. 50. 



107. Triangle abc (ab = 2 in., ac = lj^ in. and be = 2^ in.) lies 
in a plane perpendicular to H and inclined at 45° to V. The side be 
is inclined at 30° to H. Show plan and elevation of the triangle. 

108. A line making 60° with X is the plan of a line which makes 
45° with its V projection. Show the V projection. 

109. Figure 49 shows a regular tetrahedron of 2 in. edge. Find 
6 and </> for the edge E. 

110. A line has its H trace 1 in. from X and its V trace 3 in. to the 
right and 2 in. from X. Find its true length and its angles 6 and </>. 

111. Find A h when </> is 45° (Fig. 50). 

112. A ray from a point 3 in. from H and 2 in. from V is reflected 



30 



DESCRIPTIVE GEOMETRY 



from a point in V lj^ in. from H and 2 in. to the right. Where will 
the reflected ray meet H? 

113. A triangle having sides 4 in., 3 in. and 2 in. lying in H with the 
3 in. side in X, is revolved about the 4 in. side until the plane of the 
triangle is vertical. What angle does the 3 in. side then make with 
V? 

114. In the orthographic projection of shadows the direction of the 





Fig. 51. 



Fig. 52. 



x- 



/"- 



projections of the rays of light are assumed as shown in Fig. 51. What 
angle does a ray make with H and V? 

115. A ray of light is inclined at 45° to V and its plan is inclined at 

60° to X. Assume a point in the 
first quadrant 1 in. from V and 
% in. from H, and find its shadows 
A h on H and V. 

116. A cube of 3 in. edge has a 
hole drilled at a % in. deep and at 
b 1 in. deep. At what point of 
the top and at what angles with 
top and front faces must a drill 

be started to join the ends of the 

holes (Fig. 52)? 

117. A sphere which is tangent 
Fig. 53. to both H and V has its center on a 

line which meets H 2 in. from X 
and meets V 4 in. to the right and 3 in. from X. Show projections 
of the sphere. 

118. A line drawn through p intersects the lines A and B (Fig. 53). 
Find and <f> for the line. 



?/" 



jS* 



PLANES WITH REFERENCE TO H AND V 31 

119. A line is drawn from one corner to the center of a rectangular 
block 2 in. X 3J£ in. X 5 in. At what angles is this line inclined to 
the three faces of the block? 

120. The triangle abc (ab = 2 in., be = 3 in. and ac = 3^ 2 m -) 
has the point a in V, 2 J^ in. from H and 1% in. from P. b is in P, 

2 in. from H. c is in H. Show the projections of the triangle. 

121. Point d lies in V, 2 in. from H and 1J^ in. from P. Point e 
lies in H, 1 in. from V and 2 in. from P. Point /lies in P, 2}^ in. from 
d and 3 in. from e. Show H, V and P projections of the triangle def. 

122. A rectangle J^ in. X 2 in. has its short edges parallel to H. 
Its long edges are inclined at 30° to H and at 30° to V. Show H and 
V projections of the rectangle. 

123. A line 3 in. long has one end in H and the other in V. = 15°, 
<f> = 45. Show the projections of the line. 

124. A pyramid having an altitude of 1% in. has its axis in X and 
the sides of the base parallel to H and V. The face edges of the pyra- 
mid make 30° with H and 45° with V. . Show a P projection of the 
pyramid. 

125. A rectangular block 2 in. X 3 in. X 4 in. has a hole drilled 
through its center, the axis of the hole being inclined at 45° to a 

3 in. X 4 in. face and at 30° to a 2 in. X 4 in. face. 
Show where the drill pierces the faces of the 
block. 

126. A line 3 in. long has one end in H. The other 
end lies at a point 2 in. from H and 3 in. from 
V. = 30°. Show projections of the line and find 
its angle 6. 

127. The stick shown incomplete in Fig. 54 has -p r * 
the end face / against H. The other end face is cut 

to fit against P. The stick is inclined at 30° to V. Complete the 
views indicated. 

128. Show the projections of a line inclined at 30° to H and at 
60° to V. 

IX. Planes with Reference to H and V. 

A plane in space has the following distinctive data 
with reference to H and V (Fig. 55). its H trace, its V trace, 
the angle K, or true angle between the H and V traces, and 
the angles 6 and </>, or the angles made by the plane with H 
and V respectively. Any two of these are enough to 
determine the other three. 



f 


* < 


x$° 




** < 



32 



DESCRIPTIVE GEOMETRY 



Problem 16. — Given the traces of a plane, to find the angles 
of inclination with H and V (Fig, 56) . 

If a plane is tangent to a cone of revolution, the plane 




Fig. 55. 



Fig. 56. 



will make the same angle with the plane of the base that 
an element does. Hence, to find construct a cone with 
base in V and tangent to the V trace of the plane, and its 
axis in H and vertex in the H trace of the plane. The angle 
which the elements of this cone make with V will be the re- 
quired angle. Similarly to find 6 take the base of the cone 

in H tangent to the H trace, and 

the axis in V, vertex in the V trace. 
Problem 17. — Given the traces of a 

plane, to find the angle K between the 

traces (Fig. 57). 

Revolve the plane into H or V, as 

in Problem 9, and the true angle will 

appear. 

Problem 18. — Given one trace of 

a plane and the angle which the plane 
makes with the corresponding plane of projection, to find the 
other trace. 

Given 3 h and 6 (Fig. 58). Assume the base of a cone 
in H and tangent to 3 h , the axis of the cone being in V. 




Fig. 57. 



PLANES WITH REFERENCE TO H AND V 



33 



Draw an element of the cone making the given angle 6 with 
H, thus locating the vertex, through which 3 V must pass. 
There are two solutions of this problem, according as the 
vertex of the cone is taken above or below X. 

Problem 19. — Given one trace of a plane and the angle 





Fig. 58. 



Fig. 59. 



which the plane makes with the other plane of projection^ to 
find the other trace. 

Given 4 h an d <t> (Fig. 59) . From any point of 4 h draw a 
line perpendicular to X, which will represent the axis of 
a cone having its base in V. From the same point of 4 h 
draw a line making the given angle </> with X. This will 
represent an element of the same cone. Now draw the base 
of the cone, and 4 V will be tangent to it as shown. There 
are two solutions, according as 4 V is tangent above or below 
X. 

Problem 20. — Given one trace of a plane and the angle 
K, to draw the other trace. 

Given 2 h and K (Fig. 57). Lay off the given angle as 
though the plane had been revolved into H. Make the 
counter revolution, which gives the required trace. 

Problem 21. — Of a plane given the angles K and 6 or 0, 
to draw the traces. 

Given K and 8 (Fig. 60). Construct a cone with base in 



34 



DESCRIPTIVE GEOMETRY 



H, axis in V, the elements of which make the given angle 6 
with H. If the required plane is tangent to this cone, the 
element of tangency and the traces form a right triangle of 
which a side and two angles are known. Construct a right 
triangle with an element of the cone as one side, and the 
given angle K as the opposite angle. If this triangle is 
conceived to move in such a way that A remains on the sur- 
face of the cone and B remains in V, then when the point 





Fig. 60. 



Fig. 61. 



CB comes to X, C and B will coincide with the traces of the 
plane. Take the vertex of the cone as center and with the 
length of B as radius draw an arc. From the point where 
this arc cuts X draw the H trace tangent to the base of the 
cone, and the V trace through the vertex of the cone. 

Problem 22. — Given the angles 6 and </>, to draw the traces 
of the plane. 

Note.— The limits of the value of 6 + are 90° and 180°. 

A plane tangent to a 6 cone and also to a cone will 
evidently make the required angles with H and V. In 
order to locate the cones in such a way that it will be pos- 



PLANES WITH REFERENCE TO H AND V 35 

sible to draw one plane tangent to both of them, it is 
necessary that both cones be tangent to a sphere with its 
center in X. With any convenient radius draw the pro- 
jections of the sphere, Fig. 61, and draw the 6 and </> cones 
tangent to it. The H trace of the plane will pass through 
the vertex of the <£ cone tangent to the base of the 6 cone. 
The V trace will pass through the vertex of the 6 cone tan- 
gent to the base of the <£ cone. 

Exercises 

129. 2 h is perpendicular and 2 V is inclined at 60° to X. Find 
and 0. 

130. 3 h and 3 V appear as one straight line making 60° with X. 
Find and<£. 

131. K equals 75°, equals 60°. Find the traces of the plane. 

132. equals 75°, <f> equals 30°. Find the traces of the plane. 

133. The H trace of a plane is inclined at 30° to X. K is 60°. 
Find the V trace of the plane and its angles 6 and <f>. 

134. equals 30°, <£ equals 60°. Find the traces of the plane. 

135. cj> equals 45°, K equals 90°. Find the traces of the plane. 

136. <f> equals 45°, K equals 60°. Find the traces of the plane. 

137. 3 h is inclined at 30° to X. <f> equals 60°. Find 0. 

138. 4 h is inclined at 15° to X. K equals 120°. Find and 0. 

139. A corner of a cube is cut off in such a way that the face left 
makes 135° with one face of the cube and 120° with the second. What 
angle does it make with the third face? 

140. The H, V and P traces of a plane form a triangle of sides 2 in., 
2}4 in. and 3 in. What angle does the plane make with H, V and P? 

141. The oblique section of a square stick makes angles of 75° and 
45° with adjacent faces. The longer sides of the section measure 
1}^ in. Show a right section of the stick. 

142. On a square building having a hip roof the hip rafter makes an 
angle of 30° with H. What angle does the plane of the roof make 
with H? 

143. To what angle must the top of the hip rafter be beveled to lie 
in the planes of the roof? 

144. A building 20 ft. by 30 ft. has a hip roof, and the hip rafters 
are 18 ft. long. At what angles are the roof planes and the hip rafters 
inclined to the horizontal? Scale, % in. 1 ft. in. 

145. Triangle abc {ab = 2 in., be = 3 in. and ac = 4 in.) lies in a 



36 



DESCRIPTIVE GEOMETRY 



plane inclined at 45° to V. ab lies in H and ac in V. Show projec- 
tions of the triangle. 

146. Figure 62 shows the bottom portion of a bay window. A 
side face abc measures ab = 2% ft., 
ac = .5 ft. and be = 6 ft., and is inclined 
at 45° to H. Show the true form of the 
middle face. Scale, J^ in. = 1 ft. in. 

147. The edges ac and dc of the bot- 
tom portion of a bay window are in- 
clined at 30° to the wall, ed and ab 
measure 2 ft., da 3 ft., and eb 5 ft. At 
what angle is "the side face abc inclined 
to the wall (Fig. 62)? Scale, 1 in. = 1 
ft. in. 




Fig. 62. 



X. Development of Surfaces. 

Developing a surface consists of unrolling or unfolding 
it until it lies entirely in one plane. The development of 
a surface is a pattern which may be cut out of paper or sheet 
metal and be rolled or folded so as to form the surface. 

Whether or not a surface can be developed depends on the 
nature of the surface and the way it is generated. 

When a line changes its position it generates a surface. 
The nature of this surface depends upon the kind of line and 
the character of its motion. The same surface may be 
generated in more than one way, as for example the cylinder 
of revolution which may be formed by revolving one of two 
parallel straight lines about the other as axis; it may also 
be generated by moving a circle so that its center travels 
along a straight line perpendicular to its plane. 

A surface which can be generated by a straight line is 
called a ruled surface. Only ruled surfaces are developable, 
although approximate developments can be made of other 
surfaces. Not all ruled surfaces can be developed. Only 
those ruled surfaces can be developed which may be brought 
into complete coincidence with .a plane without changing 
the relative positions of any two consecutive straight line 



DEVELOPMENT OF SURFACES 37 

elements. This means that any two consecutive elements 
must be in the same plane, that is they must intersect 
or be parallel, as in the case of the cone or cylinder. 

The line of shortest distance between two points on a 
developable surface will form a straight line in the developed 
surface. 

Problem 23. — To develop the surface of a polyhedron. 

The method of development must in general consist in 
finding the true shape and size of each of the faces, and 
arranging them so that as far as desirable the same edges 
will be coincident after development as before. 

Problem 24. — To develop the lettered surface of a prism. 

Consider the face of the prism as placed against a plane 
and the prism rolled about each lateral edge, unwrapping 
its faces as it turns, till each face has come in contact with 
the plane. The lateral edges of the prism will remain 
as parallel lines of distance apart equal to the breadth of the 
corresponding faces. The total width of the development 
will be the perimeter of the prism. Any right section of the 
prism will develop into a right line perpendicular to the 
edges. If the bases of the prism are oblique the lengths of 
the edges can be set off from the development of any as- 
sumed right section. 

Problem 25. — To develop the lateral surface of a pyramid. 

If the pyramid is rolled about its edges, as in the case 
of the prism, the vertex of the pyramid rema ; ns at a point 
and the edges develop as lines radiating from this point. 
The true lengths of the various edges being set off, the 
remaining boundary lines of the development can then be 
drawn. 

A regular pyramid will develop into a series of equal 
isosceles triangles w r ith a common vertex. A truncated 
pyramid can be developed by first developing the complete 
pyramid as above, and then taking away the development of 
the portion which is cut off by the cutting plane. This 



38 DESCRIPTIVE GEOMETRY 

will consist in finding the true lengths of the edges from 
the vertex to the cutting plane, and setting off these true 
lengths in the development from the common vertex along 
the developed edges. 

Problem 26. — To develop the convex surface of a cylinder. 

The cylinder, being considered as a prism with an infinite 
number of sides, may be developed as the prism. The 
distance between the first and last element will be equal 
to the perimeter of the cylinder. A right section of the 
cylinder will develop into a right line perpendicular to the 
elements. If the cylinder has oblique ends the length of 
any element may be taken from any assumed right section 
and laid off in its developed position from the developed 
right section. Enough elements must be assumed so that 
the arc and chord between them are practically equal. A 
smooth curve drawn through the points located in the 
development will determine the outline of the development. 

Problem 27. — To develop the convex surface of a cone. 

If the cone is oblique it may be developed as a 4 pyramid, 
by assuming points in the base such that arc and chord 
drawn between them are practically the same, and con- 
sidering the elements drawn to these points as edges of a 
pyramid. 

A right cone, or cone of revolution, will develop as the 
sector of a circle having a radius equal to the slant height 
of the cone, and the arc equal in length to the circumference 
of the case. The developed angle can be calculated from 
the relation 

Radius of the base _ Developed Angle 
Slant Height 360° 

For oblique sections of cones of revolutions, elements 
are assumed, developed position found and true lengths 
laid off from the vertex. The curve of the developed base is 
drawn through the points thus found. 



DEVELOPMENT OF SURFACES 



39 



Exercises 

148. Show the developed surface of a regular tetrahedron of 2 in. 
edge. 

149. A pentagonal prism of 1 in. face is terminated at one end by a 
plane perpendicular to the axis, at the other end by a plane making 
60° with the axis, so that the two shortest edges are l}i in. long. 
Show the development of the lateral surface of the prism. 

150. A pyramid having a hexagonal base of 1 in. side has an edge 
perpendicular to the base and 2 in. long. Develop the lateral surface. 



< 2 - >i 





x- 



Fig. 63. 




151. Develop the complete surface of the pyramid shown in Fig. 
63. 

152. A vertical cylinder of 1 in. diameter is intercepted between 
two planes, one inclined at 45°, the other at 60° to H. The two 
planes meet H in the same line 2 in. from the axis of the cylinder. 
Show the development of the convex surface of the cylinder. 

153. Develop the convex surface of the cylinder which is pierced 
by a 1 in. square hole (Fig. 64). 

154. A cone of revolution, vertex angle of 60°, is cut by a plane 
making 60° with the axis of the cone and cutting the axis at a point 
2 in. from the vertex. Develop the convex surface. 

155. A cone of revolution, vertex angle of 60°, is cut by two planes, 
one inclined at 60° to the axis of the cone at a point 2 in. from the 



40 



DESCRIPTIVE GEOMETRY 



vertex, the other at 45° to the axis at a point 1 in. from the vertex. 
Develop the convex surface included between the planes. 

156. A 2 in. circle in H is the base of a cone of 2 in. altitude, the 
plan of the vertex falling J^ in. outside the base. Develop the convex 
surface. 

157. An oblique cone has for its base a 2 in. circle in V. One 
element of the cone is perpendicular to V and 3 in. long. Develop 
the convex surface of the cone. 

158. A regular pyramid has a hexagon of 1 in. side as a base, and 
an altitude of 3 in. It is cut by a plane making 45° with the axis of 

the pyramid and cutting the axis at its middle 
point. Develop the lateral surface of the truncated 
pyramid. 

159. In the pyramid of Exercise 158 show the 
projections of the shortest line that can be drawn 
on the surface between two points on opposite 
edges, one point being at a true distance of 3 in., 
the other 2 in., from the vertex. 

160. A hexagonal nut of 1 in. side is chamfered 
at an angle of 60° to the axis, to a circle 1J^ in. 
in diameter. Develop its surface. 

Fig. 65. 161. A cylinder of 1J^ in. diameter is termi- 

nated at one end by three planes forming a pyra- 
mid of 1 in. edge, as shown in Fig. 65. Develop the surface. 

162. A cord is drawn tight over a smooth cone between two points 
on opposite elements, one of the points being in the base of the cone, 
the other halfway between the base and vertex. The altitude of the 
cone is 2 in., and its vertex angle 60°. Find the true length of the 
cord between the two points, and show projections of the cone and 
cord. 




XI. Line Contained in Plane. 

If a line lies in a plane, the traces of the line lie in the 
corresponding traces of the plane. 

Problem 28. — Given one "projection of a line contained 
in a plane, to find the other projection. 

Given the plane 1 and A h (Fig. 66). This determines the 
H trace of A on l h , and the H projection of the V trace 
on X. Since the V projection of the H trace lies in X and 



LINE CONTAINED IN PLANE 



41 



the V trace lies in l v , A v which contains these points is 
readily determined. 

Problem 29. — Given one projection of a point contained 
in a plane , to find the other projection. 

Through the given projection draw the corresponding 
projection of any line of the plane, that is any line having 
its trace in the traces of the given plane. Find the other 
projection of this line by the method of Problem 28, and on 





Fig. 66. 



Fig. 67. 



this locate the required projection of the point by a per- 
pendicular to X through the given projection. 

A convenient method is to take the auxiliary line parallel 
to H or to V. If parallel to H, as in Fig. 67, the V projec- 
tion of the auxiliary line B is parallel to X and the H pro- 
jection parallel to the H trace of the plane. 

Problem 30. — To find the traces of the plane containing 
two intersecting lines or two parallel lines. 

Note. — If two lines in space intersect, the point of 
intersection of the H projections and the point of inter- 
section of the V projections, being the two projections of 
the same point, must lie in the same perpendicular to X. 

Since the traces of a line must lie in the corresponding 
traces of the containing plane, the line joining the H traces 
of the two lines will be the H trace of the plane. Similarly 
the line joining the V traces of the two lines will be the V 



42 DESCRIPTIVE GEOMETRY 

trace of the plane. The two traces of the plane must of 
course meet in X. 

Problem 31. — To find the traces of the plane containing 
a given point and a given line. 

Draw a line joining the given point with any point of 
the given line. The plane of these two lines is the one 
required, and its traces can be found by the method of the 
preceding problem. 

Problem 32. — To find the traces of the plane containing 
three given points. 

Connect the points by lines so as to form two intersecting 
lines. The plane of these lines is the one required. 

Problem 33. — To find the traces of the plane containing 
a given point and parallel to two given lines. 

Through the given point draw lines parallel to the given 
lines. The plane of these lines is the one required. 

Problem 34. — To find the traces of the plane containing a 
given line and parallel to a second given line. 

Through any point of the first given line draw a line 

parallel to the second given line. 
The plane of these two lines is 
the one required. 

Problem 35. — To find the traces 
of the plane containing a given 
point and parallel to a given 
plane. 

Through the given point draw 
a line parallel to any line of the 
given plane. The plane containing this line and having 
its traces parallel to those of the given plane is the 
one required. A convenient auxiliary line to use is one 
parallel to one of the traces of the given plane. 

In Fig. 68 p is the given point and 1 the given plane. 
Line A is drawn through p and parallel to l h . A v will there- 
fore be parallel to X. The V trace of line A is a point in 




LINE CONTAINED IN PLANE 



43 




the V trace of the required plane. Its traces are drawn 

parallel to l v and l h . 

Problem 36. — To find the lines of a given plane which 

make a given angle with H or 

V. 
Note. — The angle which a 

line of a plane makes with H 

or V cannot be greater than 

the angle which the plane 

itself makes with H or V. 
Let it be required to find 

the lines of plane 3 which 

make an angle a with H 

(Fig. 69). Take any point 

of the given plane as the 

vertex of a cone with its base in H and its elements 

making the given angle with the plane of the base. The 

points of intersec- 
tion of the circle 
of the base with 
the H trace of the 
plane will indicate 
the elements of the 
cone which coincide 
with the plane, and 
hence A and B are 
the desired lines. 

Problem 37 — 
To find the pro- 
jections of a point 
or line of a given 
plane when the 

point or line has been located in the revolved position of the 

plane. 

Suppose that the plane 2 has been revolved into H and a 




44 



DESCRIPTIVE GEOMETRY 




Fig. 71. 



point (a) located in the revolved position. Required to 

counter revolve the plane and find the projections of a. 

First Method (Fig. 70) . — Through (a) draw any line of the 

plane, such as (B) and 
locate its revolved V 
trace (v) . Counter re- 
volve this to v, and 
both projections of 
line B may be drawn, 
as we now have both 
its traces. Through 
(a) draw a perpendic- 
ular to the axis of 
revolution 2 h , which 
determines a h on B h . 
a v is located on B v 
by a perpendicular 
from a h . Lines C 
(parallel to 2 V ) and D (parallel to 2 h ) show two other lines 
either one of which may conveniently be used instead of B. 
Second Method (Fig. 71). — This method makes use of the 
right triangle construction of Problem 7, page 19. If a 
number of different points lying 
in a plane are revolved into H 
about the H trace of the plane 
as axis, then all the right tri- 
angles used in the construction 
are similar. In Fig. 71 suppose 
that plane 2 has been revolved 
into H by the right triangle 
method. Then take (a) as the 
revolved position of the point whose projections are required. 
From (a) draw a perpendicular to 2 h . Take the distance s 
and lay it off along the hypotenuse of the right triangle, 
and complete a right triangle similar to the original one. 




Fig. 72. 



LINE CONTAINED IN PLANE 



45 



The distances t and v will then locate a h and a v as shown in 
the figure. 

In general the first method is more convenient for locating 
a single point, while the second method may be found better 
when a considerable number of points are to be located. 

Exercises 

163. Show the traces of the plane AB (Fig. 72). 

164. The line B is parallel to X, % in. from H and 1 in. from V. 
The point a is J^ in. from V and 1J^ in. from H. Find the traces of 
the plane aB. 

165. Does point a lie in plane 2 (Fig. 73)? 




Xr 



tfU-r 



— IC\j| 

-y i Sr 






Fig. 73. 



Fig. 74. 



166. Show the traces of the plane abc (Fig. 74). 

167. Triangle abc (ab — 1 in., ac = 2% in. and be = 2 in.) lies in a 
plane inclined at 60° to V. ab is parallel to V and % in. from it. 
ac is parallel to H. Show the projections of the triangle. 

168. Find the traces of the plane containing the point e and parallel 
to the lines C and D (Fig. 75). 

169. Find the traces of the plane containing the line C and parallel 
to the line D (Fig. 75). 

170. Plane 3 has its H trace inclined at 45° to X and its V trace at 
30°. Find a point in the plane which is V/i in. from V and 1 in. 
from H. 

171. A rectangle 1 in. X 3 in., with the long side inclined at 30° 
to X, is the plan of a square the upper edge of which is 34 m - below H. 
Show the elevation of the square and the traces of the plane containing 
it. 



46 



DESCRIPTIVE GEOMETRY 



172. Plane 4 contains point c (Fig. 76). Find 4 V . 

173. Find a line of plane 2 of Fig. 73, the line to be inclined at 
equal angles to H and V. 

174. Find the traces of the plane cD (Fig. 77). 

175. Find the revolved positions of c and D when the plane cD is 
revolved into H about its H trace (Fig. 77). 




Fig. 75. 



Fig. 76. 



176. Point p, lying in plane 2 and % in. from V, moves to the 
position yp) when the plane is revolved into H (Fig. 78). Find 2 V 
and p v . 

177. Plane 2 has its H trace inclined at 30° to X and its V trace at 
45°. Show the projections of a line lying in the plane, parallel to H 
and 1 in. from it. 




Fig. 78. 



178. A point in plane 2 of Exercise 177, is Y 2 in. from 2 h and 1 in. 
from 2 V . Show the projections of the point. 

179. A triangle in H, sides 1% in., 1 in. and 2 in., with the 2 in. side 
parallel to X, is the plan of a triangle situated in a plane which is 
inclined at 60° with H and 45° with V. Show the elevation of the 
triangle. 



LINE CONTAINED IN PLANE 



47 



180. Find the angle K of the plane abc (Fig. 79). 

181. 5 h and 5 V are parallel to X and each 2 in. from it. Through a 
point of the plane % in. from H draw a line of the plane, the line to 
make 30° with H. 

182. Find the traces of the plane AB (Fig. 80). 



J 



.** 



x- 



ii 



1 7." 



.2" 
6 



— r 



rii 



Fig. 79. 




Fig. 80. 



183. Find the traces of the four planes of the tetrahedron of Fig. 49. 

184. Line A lies in plane 3 (Fig. 81). Show the projections of a 
second line of the plane, intersecting A at a true angle of 45°. 




Fig. 81. 



Fig. 82. 



185. Show the projections of a regular hexagon of 1 in. side lying 
in plane 3 of Fig. 75, with one side in 3 h . 

186. Point p, whose H projection is shown in Fig. 82, lies in a 
plane whose angle K = 45°. When the plane is revolved into H the 
point moves to the position (p). Find the traces of the plane and 
the V projection of the point. 

187. Point p, whose H projection is shown in Fig. 82, lies in a plane 



48 



DESCRIPTIVE GEOMETRY 



whose angle 6 = 45. When the plane is revolved into H the point 
moves to the position (p). Find the traces of the plane and the V 
projection of the point. 

188. A regular tetrahedron has edges 2J^ in. long. Find the lines 
of a face which make 60° with the plane of the base. 



b h 



x- 



*> 



a 1 



5"- 



Fig. 83. 



z 



Y 




189. A ray of light from b is reflected from V at the point a, to P 
and thence to H. Find the points of meeting P and H, and the 
angle at which the ray meets H (Fig. 83;. 

190. Find the lines of plane 4 (Fig. 84), passing through p and 
making an angle of 45° with V. 

191. A point p is 2V 2 in. from H and V. Draw through p two 




x- 




Fig. 86. 



lines each making 60° with H, the plans of the lines making 120° 
with each other. Find the angle between the lines. 

192. Plane 2 has its H trace at 30° to X and its V trace at 45°. 
A point }i m - f rom ^ and 2% in. from 2 V is the elevation of a point 



LINE CONTAINED IN PLANE 



49 



which is Y2 in. from V. Show the traces of the plane parallel to 2 and 
containing the point. 

193. Through a pass a plane whose traces are perpendicular to the 
corresponding traces of plane 2 (Fig. 73). 

194. Find a line of plane 2 of Exercise 192, which will make 45° 
with X when 2 is revolved into H. 

195. Show the projections of a line of plane 2 of Exercise 192, which 
will make the same angle with X when the plane is revolved into H 
about its H trace as it does when the plane is revolved into V about 
its V trace. 

196. Construct a square on the diagonal ab, having its plane verti- 
cal (Fig. 85). 

197. A and B (Fig. 86), are the principal axes of the H projection of 



.a h 





M 




€ h 


• 
1 
1 


y 








1 

i 

1 


A 








1 




d v 


*p> 


e v 


1 



Fig. 87. 



x4 



L 



a* 



2"- 



Fig. 88. 



b h 



-27- 



:*_i 






a circle. Show the principal axes of the V projection of the same 
circle. 

198. The point p lies in the plane def (Fig. 87). Find its H projec- 
tion without using the traces of the containing plane. 

199. Show the projections of a line parallel to V and % in. from it, 
lying in the plane abc (Fig. 88). Do not use the traces of the contain- 
ing plane. 

200. Find the projections of a line passing through b, inclined at 
60° to V and lying in the plane abc (Fig. 88). Do not use the traces 
of the containing plane. 

201. Show the projections of the bisector of the angle acb of the 
triangle of Exercise 82. 

4 



50 



DESCRIPTIVE GEOMETRY 



XII. Intersections. 

Problem 38. — To find the line of intersection of two given 
planes. 

The required line lies in both given planes, therefore 
its H trace must lie in the H traces of both planes, or at their 




Fig. 89. 

point of intersection. Similarly, the V trace of the line 
of intersection is at the point of intersection of the V traces 
of the planes. 

Case a. — When the corresponding traces intersect within 

the limits of the drawing 
(Fig. 89) . The line is readily 
determined from its traces. 

Case b. — When one pair 
of traces is parallel, the 
other intersecting (Fig. 90). 
The pair of intersecting traces 
gives one point of the re- 
quired line. The direction 
of this line will be parallel 
to the parallel traces; for if two intersecting planes in- 
tersect a third plane in parallel lines, all three inter- 
sections will be parallel. Therefore 9-10 v is drawn parallel 
to 9 V and 10\ and 9-10 h parallel to X, that is parallel to 
the H projections of 9 V and 10 v . 

Case c. — When one plane is parallel to H or V. In 




Fig. 90. 



INTERSECTIONS 



51 



•Fig. 91 plane 7 is taken parallel to V. The pair of inter- 
secting H traces gives the H trace of the required line. 
The direction of this line will be parallel to the V trace 
of the other plane; for if two parallel planes are inter- 
sected by a third plane, their 
lines of intersection are parallel. 
Case d. — When one pair of 
traces does not intersect within 
the limits of the drawing, or 
when all four traces meet X in 
the same point (Fig. 92). As in 
the previous case, one point of 
the line is determined by the 
intersecting traces. A second 
point can be found by assuming an auxiliary plane, pref- 
erably perpendicular or parallel to one of the planes of 
projection, intersecting both the given planes and cut- 
ting a line from each. These lines meet in a point com- 




Fig. 91. 




Fig. 92. 



mon to all three planes, and hence a point of the required 
line of intersection. 

Case e. — When neither pair of traces meets within the 
limits of the drawing (Fig. 93). In this case two auxiliary 
planes are taken, giving two points in the required line of 
intersection. 



52 



DESCRIPTIVE GEOMETRY 



Problem 39. — To find the point of intersection of a given 
line and a given plane. 

If any auxiliary plane is passed through the line, this 
plane will intersect the given plane in a line which contains 
the required point. It is generally most convenient 





Fig. 93. 



Fig. 94. 



to take the auxiliary plane perpendicular to either H or V, 
in which case one trace of the plane coincides with the 
corresponding projection of the line, and the other trace 
is perpendicular to X. 

In Fig. 94, 2 is the given plane, A the given line and 

the auxiliary plane 4 is 
taken perpendicular to V. 
The line of intersection of 
2 and 4 meets the given 
line A in the required 
point M. When the aux- 
iliary plane is taken perpen- 
dicular to H or V, then one 
trace of the plane and the 
FlG * 95 ' corresponding projections 

of the given line and of the line of intersection coincide. 
In the figure, A v , 4 V and ~2tf coincide, so that there- 
quired point must be located from the H projections, and 
its V projection determined by a perpendicular to X. 




, 



INTERSECTIONS 



53 



Special Cases. — If the given line is in a position nearly 
parallel to H and V, it may happen that the auxiliary 
plane will not have its traces intersecting the corresponding 
traces of the given plane within 
the limits of the drawing. In this 
case the line of intersection must 
be found by the methods of Prob- 
lem 38, case e. 

If the given line is parallel to H 
and V, the auxiliary plane will be 
parallel to one plane of projection. 
The line of intersection with the 
given plane will then be found by the method of Problem 
37, case c. This is illustrated in Fig. 95, in which A is 
the given line, 2 the given plane and 4 the auxiliary plane 
parallel to H. Notice the similarity with Fig. 67. 




Fig. 96. 



JL 



X * : 



X H 






CM 



Fig. 97. 



Fig. 98. 



Exercises 

202. Show the line of intersection of the planes 1 and 2 (Fig. 96). 

203. Show the line of intersection of the planes 3 and 4 (Fig. 97). 

204. Show the line of intersection of the planes 5 and 6 (Fig. 98). 

205. Show the line of intersection of the planes 7 and 8 (Fig. 99). 

206. Show the point of intersection of the plans 2, 3 and 4 (Figs. 
96 and 97). 

207. Show the line of intersection of planes 9 and 10 (Fig. 100). 



54 



DESCRIPTIVE GEOMETRY 



208. Find the point of intersection of planes 2, 3 and 4 (Figs. 
108, 101 and 104). 

209. Find the point of intersection of line A with plane 3 (Fig. 101). 

210. Line A is parallel to X, 1 in. from 
V and 2 in. from H. Find its point of 
intersection with plane 2 (Fig. 96). 

211. 4 h is inclined at 45°, and 5 h at 
60° to X. 6 for 4 is_30°, and for 5 is 45°. 
Find <f> for the line 45. 

212. Find the point of intersection of 
line A with plane 2 (Fig. 102). 

213. Find the point of intersection of 
line B with plane 3 (Fig. 103). 

214. Find the point of intersection of 
line C with plane 4 (Fig. 104). 




Fig. 




Fig. 102. 



Fig. 103. 



INTERSECTIONS 



55 



215. What length of line parallel to X, 1 in. from H and 2 in. from 
V is intercepted between planes 7 and 8 (Fig. 105) ? 

216. Show the projections of the point 9A (Fig. 106). 

217. The plane 7 of Fig. 105 is intersected by a line whose 
projections cross at a point J^ in. below X and 1 in. from 7 V , and are 





Fig. 104. 



Fig. 105. 



parallel to the corresponding traces of the plane. Find the point of 
intersection. 

218. The H trace of a plane which contains the point a, 1 in. from 
H and 2 in. from V, makes 75°, and the V trace 30° with X. A second 
plane also contains the point a and has its traces perpendicular to the 




X- 






Z 



Y 




a v ft* c y 



Fig. 106. 



Fig. 107. 



corresponding traces of the first plane. Find the line of intersection of 
the two planes. 

219. With the point of sight at e, show the appearance of the square 
abed on the plane P (Fig. 107). 

220. The base abc of a regular triangular pyramid is in V and of 
2 in. side, the side ab being inclined at 15° to X. Show projections of 
the vertex d when the face abd is inclined at 30° to H. 

221. A shaft A is 5 ft. above and a shaft B is 5 ft. below a floor. 



56 



DESCRIPTIVE GEOMETRY 



The distance between the centers of the shafts is 16 ft., and they are 
inclined at 30° to a vertical wall. An 8 ft. pulley on B, clearing the 
wall by 6 in., drives a 4 ft. pulley on A, on the other side of the 




■Itvi 



b" 
a" J] 



X? 



-I" A 



11 






d* 






-/" 



T? 

."Sij- 






Fig. 109. 

wall, by a belt 2 ft. wide. Show where the wall and floor are to be 
cut for the belt. Scale, % in. = 1 ft. in. 

222. Find the line of intersection of plane 2 with the plane abc 
(Fig. 108). Do not use the traces of the containing plane. 




Fig. 110. 

223. Find the line of intersection of the planes abc and def (Fig 
109). Do not use the traces of the'containing planes. 

224. Find the point of intersection of line A with plane 3 (Fig. 110). 

XIII. Perpendiculars. 

If a line is perpendicular to a plane its projections are 
perpendicular to the corresponding traces of the plane. 



PERPENDICULARS 



57 




Fig. 111. 



Problem 40. — Through a given point, to draw a line 
perpendicular to a given plane. 

Through the given projections of the point draw the 
projections of the required line per- 
pendicular to the corresponding traces 
of the given plane. 

Problem 41. — To pass a plane 
through a given point and perpendic- 
ular to a given line. 

Given point p and line A, Fig. Ill, 
to find the traces of a plane contain- 
ing p and perpendicular to A . 

The directions of the traces of the required plane will be 
perpendicular to the projections of the given line. Hence 
this problem is similar to Problem 35. Through the given 
point draw a line parallel to one of the traces. In the figure 
line B is drawn parallel to the H trace, so that its V pro- 
jection is parallel to X. The V trace of this line is a point on 

the V trace of the required 
plane, and both traces may be 
drawn in the proper directions. 
Problem 42. — To pass a plane 
through a given point and per- 
pendicular to two given planes. 

First Method. — Through the 
given point draw lines per- 
pendicular to the given planes. 
The plane of these lines is the 
one required. 
Second Method.-— Find the line 
of intersection of the two given planes. The required plane 
is perpendicular to this line of intersection, and its traces 
may be found by the method of Problem 41. 

Problem 43. — Through a given line to pass a plane per- 
pendicular to a given plane. 




Fig. 112 



58 DESCRIPTIVE GEOMETRY 

Given line A and plane 2, Fig. 112, to pass a plane through 
A and perpendicular to 2. 

From any point of the line, such as b, draw a line B 
perpendicular to the plane 2. The plane of the lines A and 
B is the required plane. 

Note. — Perpendicular planes do not have their corre- 
sponding traces perpendicular. One pair of traces may be 
perpendicular provided that one of the planes is perpen- 
dicular to that plane of projection. 

Problem 44. — Through a given point to draw a line 
perpendicular to a given line. 

First Method. — Find the plane containing the point 
and line. Revolve this plane into H or V, and with it 
the point and line. In the revolved position draw the 
required line perpendicular to the given ine. Counter 
revolve to find its projections. 

Second Method. — Through the given point pass a plane 
perpendicular to the given line. Find the 
point of intersection of the plane and line. 
The line joining the given point with 
this point of intersection is the required 
line. 

Note. — Perpendicular lines do not have 
their corresponding projections perpen- 
dicular. One pair of projections may be 
perpendicular provided one of the lines is parallel to that 
plane of projection. 

Exercises 

225. Show the point of intersection of the line passing through p 
and perpendicular to the plane 4 with the plane 4 (Fig. 113). 

226. Plane 3 has its traces appearing as one straight line at 45° to 
X. A point t in plane 3 is 1 in. from 3 h and 1J^ in. from 3 V . Show 
the H trace of a perpendicular to 3 through t. 

227. The H traces of the planes 2 and 3 form with X an equilateral 
triangle. 2 is inclined at 30° and 3 at 75° with H. Show the traces 
of a plane perpendicular to 2 and 3. 




PERPENDICULARS 



59 



228. The V trace of a plane is inclined to X at 60° and the H trace 
at 45°. Through a point in this plane which is 2 in. from H and 1 in. 
from V pass a plane which shall be perpendicular to the given plane 
and also to V. 

229. Three lines, A, B and C, meet at a point o. B makes an 





Fig. 115. 



angle of 45° with H. A is perpendicular to the plane CB. What 
angle does C make with H (Fig. 114)? 

230. Find the traces of the plane which contains point a and is 
perpendicular to planes 3 and 4 (Fig. 115). 



x- 




r i y — *p* 




Fig. 116. 



Fig. 117. 



231. A 1 in. square lies in V with sides inclined at 30° and 60° to X. 
Project the square on a plane whose traces are inclined at 45° to X. 

232. What length of line through p and perpendicular to plane 2 is 
intercepted between the planes 2 and 3 (Fig. 116)? 

233. Show the traces of the plane containing point p and perpen- 
dicular to planes 2 and 3 (Fig. 116). 

234. Show the traces of the plane containing A and perpendicular 
to 2 (Fig. 117). 



60 



DESCRIPTIVE GEOMETRY 



235. If a line and a plane are perpendicular, their angles of inclina- 
tion with H are complementary, likewise their angles of inclina- 
tion with V. Check this by finding the projections of a line making 
30° with H and 45° with V; draw a plane perpendicular to the line 
and find and <f> for the plane. 

236. A cone of revolution has its vertex at a, its base in plane 2 
and base angle 45°. Show its projections (Fig. 118). 





45r\ iff 



Fig. 119. 



237. Show the projections of the line passing through c and perpen- 
dicular to plane ^AB (Fig. 119). 

238. Find the traces of the plane passing through a, parallel to A 
and perpendicular to 2 (Figs. 118 and 119). 





Fig. 120. 



Fig, 121. 



239. Show the projections of the line through p and perpendicular 
to line A (Fig. 120). 

240. The H projection of a square having a side in H and an adja- 
cent side in V is a rectangle having sides of 1% in. and % in. Project 
this square on the plane 8 (Fig. 113). 

241. The H projection of the axis of a square base^pyramid is 
inclined at 30°, and the V projection at 45° to X. A corner of the 



ANGLES 



61 



base is in H and the adjacent edges of the base are inclined at 30° and 
60° to the H trace of the plane of the base. Show the projections of 
the pyramid. 

242. A and B are the diagonals of the base of a pyramid of 2 in. 
altitude. Its axis is perpendicular to the plane of the base. Show 
projections of the pyramid (Fig. 121). 

XIV. Angles. 

Problem 45. — To find the angle between two intersecting 
lines. 

First Method. — Find the plane of the two lines and revolve 
it with the lines into or parallel to H or V, whence the angle 
will appear in its true size. 

Second Method. — Draw a line intersecting the two lines 
and forming a triangle. Find the true length of each side 
and construct the triangle, which gives the required angle. 
In Fig. 122 let it be re- 
quired to find the angle 
between the intersecting 
lines A and B. Draw any 
third line C intersecting A 
and B. Finding the true 
length of C is simplified by 
drawing C parallel to H or 
V. In the figure it is 
drawn parallel to V, there- 
fore C v shows true length. 
Find the true lengths of 
A and B, construct the 
triangle with the three true 
lengths as sides, and the angle between A and B is the 
required angle a. 

Problem 46.— To find the angle between two given planes. 

First Method. — Pass a plane perpendicular to the line 
of intersection of the given planes. Find the lines cut 
from the given planes by this plane, and the angle between 




Fig. 122. 



62 DESCRIPTIVE GEOMETRY 

these lines, which may be found by the methods of Problem 
45, is the angle required. 

Second Method. — From any point in space draw two lines, 
one perpendicular to each plane. The angle between these 
lines, which may be found by the methods of Problem 45, is 
the supplement of the angle required. 

Problem 47. — To find the angle between a given line and 
a given plane. 

From any point of the given line draw a perpendicular 
to the plane. The angle between the given line and the 

perpendicular, which may be 
\v , found by the methods of Prob- 

^ ^^\\j* l em 44, is the complement of the 
\ / ^l\ ^v an gl e required. 

1 i \) J^> Problem 48. — To pass a plane 

\j/] ^^ through a given line and making 

Wys^^ a given angle with H or V. 

^^ Note. — The angle which the 

^ plane makes with H or V cannot 

F IG 12 3. be less than that which the line 

makes with H or V. 
Given line A, Fig. 123, to pass a plane through A making 
a given angle a with V. With any point of A as vertex 
construct a cone with base in V and base angle a. Draw 
3 V through the V trace of A and tangent to the base of the 
cone, and 3 h through the H trace of A. This gives the 
required plane. In general two solutions are possible, 
according as the V trace is drawn tangent above or below 
the base of the cone. 

Exercises 

243. What is the angle between the diagonal of a cube and an 
edge? 

244. 6 for 2 is 45°, for 3 it is 60°, for the line 23 it is 30°. Find the 
angle between the planes. 

245. Two intersecting lines A and B determine a plane. 6 for A 



ANGLES 



63 



is 45°, for B 30° and for the plane 60°. Find the angle between the 
lines. 

246. The planes 2 and 3 are at right angles to each other. 2 is 
inclined at 60° to H. The line 23 is inclined at 45° to H. What 
angle does 3 make with H? 

247. Find the angle between the planes 5 and 6 (Fig. 124j. 





Fig. 124. 



Fig. 125. 



248. Find the angles between A and AB h and between A and AB V 
(Fig. 125). 

249. Find the traces of the plane containing A and inclined at 60° 
to the plane AB (Fig. 125). 

250. A rectangular block measures 1J<^ in. X 2 in. X 3 in. What 
angle does a line drawn from a corner to the center of the block make 
with the plane determined by the three adjacent corners? 

251. A pyramid having a triangular base of 2 in. side has dihedral 
angles between its faces of 75°. What is its altitude? 

252. A plane is inclined at 60° to H 
and V. At what angle is it inclined to 
X? 

253. Find the true angle between plane 
2 and line A (Fig. 126). 

254. Line B is inclined at 30° to H, 
parallel to V and 1 in. from it. Find the 
traces of a plane containing the line and 
inclined at 45° to V. 

255. Two lines intersect at a point 1% 
in. behind V and 1 in. below H. Both 
lines are inclined at 30° to H, one at 45° and the other at 30° to V. 
Find the angle between the lines. 

256. The H trace of a plane makes 60° with X, and the plane makes 
60° with H. This plane contains two lines, one parallel to H and the 
other inclined at 45° to H. Find the angle between the lines. 




Fig. 126. 



64 



DESCRIPTIVE GEOMETRY 



Line C is per- 



Show 



r-.b h 



CVj 



X- 



^ 



n 



r- 



±.1 
a* 



r>±"_ 
L 4 



Fig. 127. 






257. Plane 4 has both traces inclined at 45° to X. 
pendicular to H. Find the angle between them. 

258. A ray of light from a is reflected from c to b (Fig. 127). 
the traces of the reflecting plane. 

259. Given the point and plane of Exercise 228, draw through the 
point a line making 30° with the plane and parallel to H. 

260. Through line A pass a plane such 
that A makes 60° with the H trace of the 
plane (Fig. 119). 

261. Show the projections of a line con- 
taining point p and intersecting line A at 
an angle of 45° (Fig. 120). 

262. Find the dihedral angles between 
the faces of the pyramid of Fig. 63. 

263. Show the traces of the plane 
which bisects the dihedral angle between 
planes 5 and 6 (Fig. 124). 

264. Show the projections of a line 
passing through a, making 60° with H and perpendicular to plane 2 
(Fig. 118). 

265. A line passing through the corner of a cube makes 75° with 
one edge and 60° with another. What angle does it make with the 
third edge? 

266. 2 h is inclined at 45° and 2 V at 30° to X. The line of intersec- 
tion of the planes 2 and 8 bisects the angle K for the plane 2. The 
plane 8 is inclined at 30° to H. Show its traces. 

267. The traces of a plane make 45° with X. Find the traces of a 
second plane which is inclined at 60° to H and is perpendicular to the 
first. Find also the line of intersection of the two planes. 

268. Find the traces of a plane containing line A and inclined at 
60° to H (Fig. 119). 



XV. Distances. 



Problem 49. — To find the distance from a given point 
to a given plane. 

Through the given point draw a perpendicular to the 
given plane and find the point of intersection of the perpen- 
dicular with the plane. Find the distance between this 
point and the given point as in Problem 10. 



DISTANCES 65 

Problem 50. — To find the distance from a given point 
to a given line. 

Draw a line from the given point perpendicular to the 
given line, as in Problem 44. The true length of this line 
is the distance required. 

Problem 51. — To find the distance between two parallel 
planes. 

First Method. — Construct a third plane perpendicular 
to the two given planes, and find their lines of intersection. 
Revolve the third plane with the lines of intersection into 
H or V, and the distance between the lines is the required 
distance. For convenience take the third plane perpen- 
dicular either to H or V. 

Second Method. — Draw a line perpendicular to the given 
planes, and find its point of intersection with each. The 
distance between these points is the required distance. 

Problem 52. — To find the distance between two parallel lines. 

First Method. — Find the plane of the two lines and 
revolve the plane, and with it the lines, into H or V. 
The distance between the revolved lines is the required 
distance. 

Second Method. — Pass a plane perpendicular to the 
given lines, and find its point of intersection with each. 
The true distance between these points is the required 
distance. 

Problem 53. — To find the shortest distance between two 
lines not in the same plane. 

Pass a plane through one of the lines and parallel to the 
other. From any point of the other line draw a perpen- 
dicular to the plane. The true length of this perpendicular 
is the required distance. 

Exercises 

269. What is the distance from a corner of a 2 in. cube to the plane 
of the three adjacent corners? 

270. Find the distance from a to B (Fig. 128). 



66 



DESCRIPTIVE GEOMETRY 



271. Find the shortest distance between the lines A and B (Fig. 
129). 

272. 2 h and 3 h are parallel, 1% in. apart and inclined at 45° to X. 
The V traces are also parallel and 1 in. apart. What is the distance 
between the planes? 

273. Find the distance from point a to plane 2 (Fig. 130). 





Fig. 128. 



Fig. 129. 



274. Find the distance between planes 3 and 4 (Fig. 131). 

275. The H projections of two points are 1 in. apart, and the V 
projections are 2 in. apart. What is the greatest and what the least 
distance apart that the points can be placed? 

276. Find the distance between lines A and B (Fig. 119). 





Fig. 130. 



Fig. 131. 



277. A line 3 in. long between its H and V traces has its H and P 
traces at the same point. The line makes an angle of 45° with H and 
15° with V. Find a point of the P plane which is 2J^ in. from both 
ends of the line. 

278. A triangular pyramid has a base of 3 in. side and an altitude 



TANGENT PLANES 



67 






,a* 



•/"- 






Li 



C A t 



/"- 



7 



$■ 



of 4 in. Show the projections of a point which shall be respectively 
K kd-j M in. and 1 in. from the three faces of the pyramid. 

279. Find the projections of a line parallel to X which is equally 
distant from the points a, b and c ^ 
(Fig. 132). 

280. The line A is inclined at 30° 
to H and 45° to V. The line B is in- 
clined at 45° to H and 30° to V. The 
shortest distance between the two 
lines is 2 in. Show their projections. 

281. Point a is 1J^ in. from V and 
}/2 in. from H. Point b is 2 in. to the 
right, \i in. from V and 1)4 in. from 
H. Plane 3 contains point a and has 
its H trace at 45° and its V trace at 
30° to X. Find the distance from 
the point b to the plane. 

282. Find the distance from point 
cto plane AB (Fig. 119). 

283. Point p is at a true distance of 1 in. from plane #~(Fig. 133). 
Find 2 V . 

284. Show the traces of a plane parallel to plane 2 and at a distance 
of 1 in. from it (Fig. 130). 

285. Line A is parallel to V and \y 2 in. from it, and inclined at 30° 

to H. Point b is lji in. from V 
and % in. from H, b v being 1 in. 
from A v . Show projections of the 
path of b when it is revolved about 
A as axis. 

286. Find the distance from 
point d to the plane of a, b and c, 
Fig. 109, without using the traces 

of the plane abc 

287. A cube of 2% in. edge has 
An edge of the cube has one end 3^ in- and 

the other 1% in. from H. Show projections of the cube. 



[Fig. 132.3 




Fig. 133. 
its center 2 in. from H. 



XVI. Tangent Planes. 

If a straight line is tangent to a curve, two consecutive 
points on the curve lie in the straight line, the two points 
being infinitely close together. Similarly if a line is tangent 



68 



DESCRIPTIVE GEOMETRY 



to a surface, it has two consecutive points in common 
with the surface. These two points, being infinitely 
close together, constitute what is commonly referred to as 
the point of tangency. 

If two lines are tangent to a surface at the same point, 
the plane of the lines is tangent to the surface, and the 
plane has three points in common with the surface. These 
points, being infinitely close together, constitute what is 
commonly referred to as the point of tangency. 

If the surface is a ruled surface, the tangent plane 

will contain two consecu- 
tive elements of the sur- 
face, and these elements 
being infinitely close to- 
gether constitute the line 
of tangency. In the case 
of doubly ruled surfaces of 
which there are but two, 
the hyperboloid and the 
hyperbolic paraboloid, the 
tangent plane at a given 
point of the surface is determined by the two elements 
through the point. 

Problem 54. — Given one projection of a point on the surface 
of a cone, to pass a plane tangent to the cone and containing 
the point. 

Consider the general case of an oblique cone, and place 
its base in V, as shown in Fig. 134. Let a h be the given 
projection of a point on the surface of the cone. Draw 
the projections of the element containing this point, as 
B h and B\ The V trace of the required plane, S v is tangent 
to the base of the cone at the V trace of B, and 3 h passes 
through the H trace of B. If the H trace of B should fall 
without the limits of the drawing or be inconvenient to 
find, any auxiliary line drawn through the vertex of the 




Fig. 134. 



TAXGEXT PLAXES 69 

cone and a point of the V trace of the plane may be drawn, 
and its H trace used instead. 

Problem 55. — To pass a plane tangent to a given cone 
and containing a given point outside its surface. 

Suppose the base of the cone assumed in H. The tangent 
plane will contain the line drawn through the vertex of 
the cone and the given point. Hence the H trace of the 
tangent plane will pass through the H trace of this line and 
be tangent to the base of the cone. The V trace will 
pass through the V trace of the line. If this point cannot 
be found conveniently, an auxiliary line may be used 
as in Problem 54. 

Problem 56. — To pass a plane tangent to a given cone 
and parallel to a given line. 

A line through the vertex of the cone and parallel to the 
given line determines the required plane as in Problem 55. 

Note. — The methods of the preceding problems apply 
equally well to oblique cones and to cones of revolution. 

Problem 57. — To pass a plane tangent to a given cylinder. 

Considering the cylinder as a special case of the cone, 
that is having its vertex at infinity, the methods of the 
three preceding problems will apply as well to the corre- 
sponding cases of the cylinder. 

Problem 58. — To pass a plane tangent to a given surface 
of revolution and containing a given point on that surface. 

This problem may be solved by a number of different 
methods and the choice of method depends on the nature of 
the surface. A perfectly general method is to construct 
a cone of revolution tangent to the surface in a circle 
passing through the given point. The required plane will 
be tangent to the cone at the element passing through the 
given point. 

Another method which does not completely determine 
the tangent plane, but is helpful in some cases, is to pass 
a plane through the given point perpendicular to the 



70 DESCRIPTIVE GEOMETRY 

axis of the surface of revolution. This plane cuts a circle 
from the surface and the given point lies on the circum- 
ference of this circle. In the plane of the circle draw a line 
tangent to the circle at the given point. The tangent 
plane must contain this tangent line. Special methods 
may be used in special cases. Planes tangent to cones of 
revolution can be found as in Problems 54, 55 and 56; 
to cylinders of revolution as in Problem 57. 

Problem 59. — To pass a plane tangent to a given sphere 
at a given point on its surface. 

Draw the radius of the sphere to the given point. The 
required plane is perpendicular to this line and contains 
the given point. 

Problem 60= — To pass a plane through a given line and 
tangent to a given sphere. 

First Method. — Pass a plane through the center of the 
sphere and perpendicular to the given line. This will cut 
a great circle from the sphere and a point from the given 
line. Revolve the plane into H or V, and locate the point 
and circle. From the point draw a tangent to the circle. 
Counter revolve and find the projections of the tangent 
line. The plane of this line and the given line is the one 
required. 

Second Method. — From two points on the given line 
construct cones tangent to the sphere. The intersection 
of the circles of contact will determine the point of tangency 
of the required plane. If the axis of one of the tangent 
cones is taken parallel to H and the other parallel to V, but 
one ellipse will be required in the construction. 

Problem 61. — To pass a plane tangent to an oblique cone 
and making a given angle with the plane of the base. 

Construct a cone of revolution with its vertex coinciding 
with the vertex of the given cone and its elements making 
the given angle with the plane of the base. The required 
plane is tangent to both cones. 



TANGENT PLANES 71 

Problem 62. — To pass a plane tangent to an oblique 
cylinder and making a given angle with the plane of the base. 

Construct a cone of revolution at any convenient place,, 
having its base parallel to the base of the cylinder and its 
elements making the given angle with the plane of the base. 
Draw a line through the vertex parallel to an element of the 
given cylinder. The plane containing this line and tangent 
to the base of the cone will be parallel to the required plane, 
which will be found by taking its traces tangent to the 
corresponding traces of the cylinder and parallel to the 
traces of the plane found. 

Exercises 

288. A cone of 2 in. base and 1^2 i n - altitude lies in contact with 
both H and V. Show the lines of contact. 

289. A cone having a vertex angle of 45° has its axis in X. Show 
the traces of a plane tangent to the cone at a point on its surface 
}4 in- from H and 3 in. from the vertex. 

290. Find the traces of the plane containing the points a and b and 
inclined at 75° to H (Fig. 135). 

291. A cone of revolution of vertex angle 60° has a slant height of 
2 in. Its base lies in V, with center 11^ in. be- 
low H. Show the traces of the plane tangent ^ 
to the cone at a point on its surface % m - from 
V and 2 in. from H. w ah 

292. An oblique cylinder has its axis inclined /\ ot^ , tf ' 
at 45° to H and parallel to V. Its base is a 1% 
in. circle in H. Show the traces of a plane 
making 60° with H and tangent to the cylinder. b v \ 

293. A sphere of 2 in. diameter, center in H, Fig. 135. 
% in. from V, is tangent to a plane which is 

inclined at 45° to II and 60° to V. Show projections of the point 
of tangency. 

294. A cone which has a vertex angle of 60° has two planes tangent 
to it at elements which make 30° with each other. What is the angle 
between the planes? 

295. A sphere 2 in. in diameter, tangent to both H and V, has a 
great circle cut from it by a plane whose traces meet X at 45°. Show 
the projections of a line tangent to this circle, the H projection of the 
line being inclined at 30° to X. 



u 



72 



DESCRIPTIVE GEOMETRY 



296. Show the projections of a cone of revolution tangent to H, V 
and P. Find also its vertex angle. 

297. An anchor ring with its axis perpendicular to H has an outside 
diameter of 3 in. and an inside diameter of 1 in. A plane tangent to 
it has its H trace inclined at 45° to X. <£ equals 60°. Find the point 
of tangency. 

298. A sphere 3 in. in diameter has its center in H, 1 in. from V. 
Show the traces of a plane tangent to the sphere at a point 2 in. from 
V and }/2 in. from H. 

299. Two spheres of 1 J^ in. and % in. diameter are tangent to both 
H and V and to each other. Show their projections. 





Fig. 136. 



Fig. 137. 



300. Show the traces of a plane tangent *to both spheres of Exercise 
299, and having = </>. 

301. The line A, Fig. 136, is the axis of a cylinder which is tangent 
to the cone. What is the diameter of the cylinder? 

302. Find 3 V if the plane 3 is to be tangent to the sphere (Fig. 137) 



XVII. Shades and Shadows. 

In architectural drawings, and occasionally in other types 
of drawings, much is added in the way of clearness and 
beauty by showing shades and shadows, and a single ortho- 
graphic projection is brought into relief, giving a much 
clearer idea of the depth of the object in a direction per- 
pendicular to the plane of the drawing. 

The direction of the rays of light may be taken in any 
way desired ? but it is almost universal practice to assume 



SHADES AND SHADOWS 73 

a direction such that both projections of a ray make 45° 
with X and from over the left shoulder. 

In the case of an actual object illuminated by light rays 
in a certain direction, there are always gradations from 
extreme high light of illumination to extreme depth of 
shadow, and further variations are introduced by reflected 
and diffused light. In mechanical drawings no such 
gradations are attempted — a surface is either light or 
dark according as rays of light do or do not reach it. 

Surfaces which are on the dark side of an object, from 
which the light is excluded by the object itself, are said 
to be in shade. Surfaces which are turned in a direction 
to receive the light, but from which the light is excluded 
by other objects or other parts of the same object, are said 
to be in shadow. On an actual object shadows generally 
appear darker than shade, but again no such distinction 
is attempted in mechanical drawings. 

When a solid body is interposed between a plane and 
the source of light, the body casts a shadow on the plane. 
Light is excluded from the space between the body and its 
shadow, and this space is called umbra, or invisible shadow, 
or shadow in space. The outline of the shadow will be 
determined by a series of light rays tangent to the solid. 
In other words the shadow outline is the shadow of the 
linear outline along which light rays are tangent to the 
solid. This linear outline on the body, which may be plane 
or otherwise, is called the line of shade, and it is evident 
that it is the line on the body separating light from shade. 
Determining the shadow of an object then reduces to the 
problem of finding the shadow of the line of shade. Since 
this line of shade consists of a series of points, any problem 
of finding shadows is merely a problem of finding shadows 
of points. 

The shadow cast on a surface by a point is the point 
of intersection of a light ray through the point with the 



74 DESCRIPTIVE GEOMETRY 

surface. The invisible shadow of a point is a straight 
line coinciding with this ray of light. 

The shadow of a line consists of the shadows of its indi- 
vidual points. In general the shadow of a line is a line, and 
its invisible shadow is a surface. If the line is straight 
its invisible shadow is a plane, and its shadow is the line 
of intersection of the plane with the surface on which the 
shadow is cast. 

The following principles should be kept in mind in plot- 
ting shades and shadows, as their application simplifies 
the work to a considerable extent : 

1. The shadow of a straight line on a plane is a straight 
line. 

2. If any line, either straight or plane curve, casts a 
shadow on a plane to which it is parallel, its shadow is 
the same length and shape as the line. 

3. If a straight line is perpendicular to H or V, then the 
H or V projection of its shadow on any surface is a straight 
line at 45° to X. 

4. The point where the shadow of one line crosses another 
line not in the same plane with it can be found by finding 
shadows of both lines on any convenient plane. From the 
point of intersection of the two shadows draw a ray of 
light backward, and where this ray cuts the second line 
is the required point. If this ray of light is produced back- 
ward till it intersects the first line, it gives the point whose 
shadow falls on the second line. 

5. The shadow of a curved line on a plane, or of any line 
on a curved surface, is determined by finding the shadows 
of a number of points of the line on the surface and joining 
the point shadows thus found. The points must be taken 
sufficiently close together to determine the required shadow 
with the desired degree of accuracy. 

6. The general method of finding the shadow of an object 
will be first, determine the line of shade; second find its 



SHADES AND SHADOWS 



75 



V.S.R 



shadow. It is sometimes more convenient to reverse this 
procedure, first finding the shadow and then by drawing 
rays of light back from the shadow, determining the line 
of shade. 

Projecting Planes and Shadow Planes 

The object is considered as being in the third quadrant, 
and plan and elevation are drawn. The planes on which 
shadows are cast are taken parallel to H and V. The hori- 
zontal shadow plane is taken as the plane on which the 
object rests, and is called the ground plane, and will be 
referred to as G. P. The ver- 
tical shadow plane is taken 
behind the object at a distance 
such that part of the shadow 
will fall on it. This plane 
will be referred to as V. S. P. 

Only that part of the 
shadow on V. S. P. which 
comes above G. P. need be 
found, and also only that part 
on G. P. which comes in front 
of V. S. P. It is sometimes 
necessary to find a point or 
two behind V. S. P. or below G. P. in order to determine 
the shadow. The two shadows will join along the line of 
intersection of the two shadow planes. 

In Fig. 138 are shown the shadows of a rectangular block. 
The method of locating the points is shown by the construc- 
tion lines. The hidden port : on of the shadow on V. S. P. 
may be shown by dotted lines as indicated, or may be 
omitted entirely. The shades and shadow of cylinder are 
shown in Fig. 139 and of a cone in Fig. 140. The lines of 
shade of the cone are best determined by first finding the 
shadow. The shadow of a cone on the plane ofits base is 




Fig. 138. 



76 



DESCRIPTIVE GEOMETRY 



determined by finding the shadow of its vertex and drawing 
tangent lines from this point to the base. The points of 
tangency determine the elements which are the lines of 
shade. 

For finding the lines of shade on a solid of revolution 
with its axis vertical, and its shadow on the G. P. the follow- 
ing method is useful. Cut the solid by a series of planes 
perpendicular to the axis of the solid, cutting the solid in 





Fig. 139. 



Fig. 140. 



circles. Find the shadows of these circles on the ground 
plane (see principle 2). The curve enveloping these circular 
shadows is the required shadow. The line of shade can 
be found by drawing light rays from the points where the 
circular shadows are tangent to the envelope back to the 
original plane sections, each section giving two points on the 
line of shade. 

Pilet's Method. This is a method of finding shadows 
of circles on surfaces of revolution, and is useful in a number 
of problems. It depends on two principles, one of which 



SHADES AND SHADOWS 



77 



has already been stated as Principle 4. The other is that 
the shadow of a horizontal circle on a vertical plane inclined 
at 45° to V shows as a circle in the V projection, the diameter 
of the shadow projection being to the diameter of the circle 
as the half diagonal of a square is to a side of the square. 
This is il'ustrated in Fig. 141, in which abed is a square 
circumscribed about the horizontal circle, and a'b'e'd' is the 
V projection of its shadow on the plane 3. 

To illustrate the application of these principles in Pilet's 





Fig. 141. 



Fig. 142. 






method, let us consider a cylindrical column with a circular 
lintel, as in Fig. 142, required to find the shadow of the 
lintel on the column. The lower circle ab is the line of 
shade. Assume a vertical 45° plane through the common 
axis of column and lintel, and find the V projection of the 
shadow of circle ab on this plane. This will be a circle with 
its center on a v b v and with diameter equal to the semi- 
diagonal of a square of side ab. Take any horizontal sec- 
tion of the column, as cd, and find the V projection of the 
shadow of this circle on the vertical plane. This will be a 
circle with center on c v d v . The intersection of the two 



78 DESCRIPTIVE GEOMETRY 

shadow circles gives two points e and /, which must be the 
shadows of the points in which the shadow of the circle ab 
cuts the circle cd. In other words, e and / are the shadows 
on the 45° plane of two points on the required curve. 
Draw light rays back, cutting c v d v in g and h, which are two 
points on the required curve. By taking a number of 
planes such as cd, any desired number of points can be 
determined and the required curve obtained. 

For further methods in shades and shadows the student 
is referred to a series of articles by Prof. A. D. F. Hamlin, in 
the American Architect, beginning in 1889. These articles 
constitute a most excellent treatise on the subject, and from 
them was taken the Pilet's method explained above. 

Exercises 

303. Find shadow of lintel on column (Fig. 143). 

304. Find shade and shadow of lintel on column (Fig. 144). 

305. Find shade and shadow of lintel on column (Fig. 145). 

306. Find shades and shadow of lintel on column (Fig. 146). 

307. Find shade and shadow of lintel on column (Fig. 147). 

308. Find shades and shadow of lintel on column (Fig. 148). 

309. Find shades and shadows (Fig. 149). V. S. P. as shown. 

310. Find shades and shadows (Fig. 150). V. S. P. as shown. . 

311. Find shades and shadows (Fig. 151). No V. S. P. 

312. Find shades and shadows (Fig. 152). V. S. P. as shown. 

313. Find shades and shadows (Fig. 153^. No V. S. P. 

314. A cylinder has as its base a 2 in. circle the plane of which is 
parallel to H. The cylinder is inclined at 60° to H and is parallel 
to V. Find shades and shadows. No V. S. P. 

315. Find shades and shadows (Fig. 154). No V. S. P. 

316. Find shades and shadows (Fig. 155). No V. S. P. 

317. Find shades and shadows (Fig. 156). No V. S. P. Each 
block is 2 in. X 1 in. X M m -, an d the upper one is inclined at 30° 
to H. 

318. Find shades and shadows (Fig. 157). V. S. P. as shown. 

319. Find shades and shadows (Fig. 158). No V. S. P. 

320. Find shades and shadows (Fig. 159). V. S. P. as shown. 

321. Find shades and shadows (Fig. 160). No V. S. P. 

322. Find shade and shadow on the ground plane of a 2 in. sphere. 



SHADES AND SHADOWS 



79 




Ik 


*-/*-» 






! 












; 


— 


\l 






1 










c 





1 1 


gj> 


"cm 

1 








G.P 


x 


f 




Fig. 149. 



Fig. 150. 

Figs. 143 to 151. 



Fig. 151. 



80 



DESCRIPTIVE GEOMETRY 




Fig. 154. 



Fig. 155. 



Figs. 152 to 155. 



SHADES AND SHADOWS 



81 




Fig. 156. 



V.S.P 




x— = 



* $ 



!" 



2" ^ 



Fig. 157. 



G.P 



x- 



-/"- 











A ' 


[ 


J 




/ v^ 


^ICVj 








i 






A G.P 




Fig. 158. 



Fig. 159. 



82 



DESCRIPTIVE GEOMETRY 




Fig. 160. 



Fig. 162. 




SECTIONS 83 

323. Find shades and shadow on the ground plane of the solid of 
revolution shown in Fig. 161. 

324. Show the projections of the shadow on plane 4 of the sphere 
shown in Fig. 162, also the shade on the sphere. 

325. Show projection of the light spot in the dome (Fig. 163). 

326. A luminous point 1 in. above the vertex of a cone is 1% in. 
from the axis of the cone. The cone has a base 2 in. in diameter and 
an altitude of 2 in. Find the lines of shade on the cone and its shadow 
on the plane of the base. 

327. With a as source of light, find shades and shadows in Fig. 164. 
No V. S. P. 

XVIII. Sections. 

When a solid is divided by a plane the surface of division 
is called a section. If in a prism the cutting plane contains 
an edge, in a cylinder an element or in a cone the axis, the 
section is called longitudinal. In a solid of revolution the 
longitudinal section becomes the meridian section. Since 
the cutting plane contains the axis and a generating element, 
all meridian sections of a solid of revolution are equal. All 
planes perpendicular to the axis will cut circles called paral- 
lels, of which the smallest, if greater than zero, is called the 
circle of the gorge, and the largest the circle of the equator. 

The general method of finding the section by an oblique 
plane is to cut both the solid and the plane by auxiliary 
planes so assumed as to cut the simplest lines, that is the 
lines most easily found, from each. Since the required 
line of intersection is common to both the secant plane and 
the surface of the solid, the lines cut from them by an aux- 
iliary plane will intersect in points of the required line. 

In order to get the simplest lines the auxiliary planes 
should in general be taken as follows : 

(a) In solids bounded by plane surfaces, coincident 
with the surfaces or through edges. 

(6) In oblique cylinders, parallel to the axis. 

(c) In oblique cones, passing through the vertex. 

(d) In solids of revolution, perpendicular to the axis. 



84 



DESCRIPTIVE GEOMETRY 



Problem 63. — To show the projections and true form of 
the section of a polyhedron. 

The section is a right lined figure which can be deter- 
mined by finding the vertices from the intersections of the 
edges with the cutting plane, or the sides of the figure can 
be determined by finding the intersections of the planes of 
the solid with the cutting plane. 

The true form of the section can be found by revolving the 
plane of the section into H or V. 

In case the position of the solid can be assumed so as to 




Fig. 165. 



bring the cutting plane perpendicular to H or V the work 
is somewhat simplified. One projection of the required 
section is then a straight line coinciding with the oblique 
trace of the cutting plane. From this the vertices of the 
other projection can readily be projected. 

Problem 64. — To show the projections and true form of the 
section of a cylinder. 



SECTIONS 85 

The projections of the section are found by drawing 
elements of the cylinder and finding their points of inter- 
section with the cutting plane. Enough points should be 
found to properly determine the outline of the section, 
which will be a smooth curve joining these points. 

In the case of a cylinder of revolution the section, not 
considering the bases of the cylinder, will be two parallel 
lines, a circle or an ellipse, according as the cutting plane 
is parallel, perpendicular or oblique to the axis of the 
cylinder. The projections of the section may befound by 
drawing a number of auxiliary planes perpendicular to the 
axis of the cylinder, as illustrated in Fig. 165. Let it be 
required to find the section of the cylinder of revolution by 
plane 3. The axis of the cylinder being vertical, a series 
of horizontal planes is drawn, each plane giving two points 
on the required curve. 

The true form of the section is found by revolving 3 into 
H about 3 h as axis. If the true form only is required, the 
lengths of the major and minor axes may be determined and 
the ellipse constructed by any convenient method. 

Problem 65. — To show the projections and true form of the 
section of a cone. 

The method of finding the projections and true form of 
the section will be the same as in the case of the cylinder. 

In the case of a cone of revolution the section, not con- 
sidering the base ; is bounded by two intersecting straight 
lines if the cutting plane passes through the vertex. Other- 
wise the section is a circle if the cutting plane is perpen- 
dicular to the axis of the cone, an ellipse if oblique to the 
axis but making a greater angle with it than the elements of 
the cone, a parabola if the same angle as the elements, and 
an hyperbola if a less angle. 

Problem 66. — To show the projections and true form of 
the section of a solid of revolution. 

Place the solid with its axis perpendicular to H or V. 



86 



DESCRIPTIVE GEOMETRY 



Cut it and the secant plane by auxiliary planes taken 
perpendicular to the axis of the solid, cutting circles from 
the solid and straight lines from the plane, which intersect 
in points of the required line. The true form is found as 
before. This method is the one used in Fig. 165, and may 
be applied to any solid of revolution. 

Exercises 



328. A hexagonal prism of 1 in. side and 3 in. long has a hole % in. 
in diameter, the axis of which coincides with the axis of the prism. 
Show the section by a plane making 60° with the axis of the prism. 

329. A pentagonal prism of 1 in. side and 4 in. long is cut by a plane 
bisecting the axis and making 45° with it. Show true form of section. 

330. Show the true form of the sec- 
tion of a cube by a plane passing 
through its center and perpendicular 
to a diagonal. 

331. A cone having a vertex angle 
of 90° has a parabola cut from it by a 
plane which is Y± in. from the vertex 
of the cone. Show the true form of 
the section. 

332. A cone of 3 in. altitude and 
base 2 in. in diameter has an hyper- 
bola cut from it by a plane % in. from 
the axis of the cone. Show the true 
form of the section. 

333. A cylinder of 2 in. diameter 
with its axis vertical is cut by a 

plane whose traces are inclined at 30° to X. Show projections and 
true form of the section. 

334. A cone of revolution has a vertex angle of 60°. Show the 
traces of a plane which will cut the cone in an ellipse whose major axis 
is 2 in. and minor axis 1 in. 

Note — See text on page 101. 

335. A cone of revolution of vertex angle 60° is cut by a plane at 
60° to its axis, 1}4 m - from the vertex. Show the projections of a 
cylinder of revolution of which the ellipse is an oblique section. 

336. The axis of a cylinder of 1J^ in. diameter is inclined at 45° to 




Fig. 166. 



SECTIONS 



87 




Fig. 167. 



H and 30° to V. The cylinder is tangent to X. Show its intersec- 
tions with H and V. 

337. A sphere of 2 in. diameter tangent to both H and V is cut 
by a plane whose traces meet X at 45° and pass through the pro- 
jections of the center. Show projections and true form of the 
section. 

338. Show projections and true form of the section of the torus cut 
by planed (Fig. 166). 

339. A solid of revolution is generated by revolving a 2 in. circle 
about a cord J^ in. from the center. Show true 
form of section cut by a plane passing through 
the center of the equator and inclined at 45° 
to the axis of the solid. 

340. A cone of vertex angle of 60° has its ver- 
tex 1 in. from H and its axis inclined at 45° to 
H. Show its intersection with H. 

341. A stick of octagonal section, the cir- 
cumscribing circle being 13^ in. in diameter, 
has its axis perpendicular to V. It is cut by a vertical plane making 
45° with V. A second stick with its axis parallel to V and inclined 
at 30° to H exactly meets the section of the first stick. Show the 
right section of the second stick. 

342. A stick, the section of which is a parallelogram, has its V trace 

and the projection of an edge shown 
in Fig. 167. What is the angle be- 
tween the two faces meeting in 
edge E? 

343. Show the section of a reg- 
ular tetrahedron cut by a plane 
passing through its center and par- 
allel to two adjacent edges. 

344. A solid of revolution is 
formed by revolving a circular arc 
of 90° about its chord which is 3 
in. long. Show the true form of 
the section of the solid by a plane 
parallel to the axis and J4 in. 
from it. 

345. Show 




Fig. 168. 



the true form of the 
section cut from the torus of Fig. 168 by the plane 2. 
Note. — This section is the Lemniscate of Bernouilli. 
346. A cone of revolution of vertex angle 60° with its vertex lying 




88 DESCRIPTIVE GEOMETRY 

in H, 1J^ in. from V, and its axis perpendicular to H, is cut by a plane 
whose H trace makes 45° with X and V trace 30°. The V trace of the 

plane is 1}^ m - from the V projection 
of the vertex. Show projections and 
true form of the section. 

347. The stick shown in plan and 
end elevation in Fig. 169 is cut by a 
Fig. 169. plane such that the plan of the section 

is an equilateral triangle. Show the true form of the section. 

XIX. Intersection of Surfaces. 

If a given surface made up of planes intersect a second 
surface the line of intersection will be made up of parts of 
the sections cut by the planes making up the first surface, 
and may be determined by the methods given under 
Sections. 

In general, to find the line of intersection of two sur- 
faces auxiliary planes are taken cutting lines from the 
given surfaces which intersect in points of the required 
line. The amount of work required to find any desired 
line of intersection will depend largely on the judgment 
used in selecting auxiliary planes so as to cut simple 
lines from the given surfaces. 

Problem 67. — To find the projections of the line of inter- 
section of two cylinders. 

Take auxiliary planes parallel to the axes of both cylinders, 
cutting straight line elements from each. These elements 
intersect in points of the required curve. 

In Fig. 170, given two oblique cylinders whose. axes 
are the lines A and B, the base of one being taken in H, 
the other in V. The first step is to determine the direc- 
tion of the auxiliary planes. Through any point in space, 
such as 6, draw lines A± and Bi parallel to the axes of the 
given cylinders, and find the plane of these lines, plane 3. 
Any auxiliary plane parallel to 3 will cut straight line ele- 
ments from both cylinders, and these elements intersect 



INTERSECTION OF SURFACES 



89 



in four points of the required line. One such auxiliary 
plane is shown as plane 4) cutting lines C and D from one 
cylinder and E and F from the other. These lines intersect 
in points a, &, c and d, which are points on the required 
curve. A sufficient number of such planes must be used 
to give points enough to determine the curve. 

The foregoing method is perfectly general, and is much 




Fig. 170. 

simpler in the case of cylinders of revolution. In this case 
one cylinder can be placed with its axis perpendicular to V, 
the other parallel to H, and the auxiliary planes are taken 
parallel to H. 

Problem 68. — To draw the projections of the line of inter- 
sections of a cylinder and a cone. 

The general method is the same as in the preceding 
problem. The auxiliary planes, in order to cut straight 
lines from both cylinder and cone, must pass through the 



90 



DESCRIPTIVE GEOMETRY 



vertex of the cone and be parallel to the axis of the cylinder. 
Hence if a line parallel to the axis of the cylinder is drawn 
through the vertex of the cone, all auxiliary planes must 
have their traces containing the corresponding traces of 
this line. 

Problem 69. — To draw the projections of the line of inter- 
section of two cones. 

Again the same general method is used, the auxiliary 
planes passing through the vertices of both cones. Hence, 
if a line is drawn joining the two vertices, all auxiliary 

planes must have their traces 
containing the corresponding 
traces of this line. 

Problem 70. — To draw the 
projections of the line of inter- 
section of a sphere and a polyhe- 
dron. 

A series of planes parallel to 
H or V will cut circles from the 
sphere and polygons from the 
polyhedron, which will intersect 
in points of the required line of 
intersection. 

Problem 71. — To draw the 
projections of the line of inter- 
section of ttvo surfaces of revolution, axes intersecting. 

Place the surfaces with axes parallel to V and one of the 
axes perpendicular to H. With center at the intersection 
of the axes take a series of spheres which intersect the given 
surfaces in circles, the elevations of which are straight lines. 
The plan of one is a circle and of the other an ellipse, but 
the points of intersection can be determined without the use 
of the ellipse. 

This method as applied to the case of a cylinder and cone 
of revolution is illustrated in Fig. 171. With center at e 




INTERSECTION OF SURFACES 



91 



the point of intersection of the two axes, draw a sphere which 
cuts both cylinder and cone in circles. The V projection 
of both circles are straight lines, the H projection of one is 
the circle shown, and of the other an ellipse, which is not 
shown. The points a, b, c and d are determined from the V 
projections and the circular H projection. A sufficient 
number of spheres must be taken to give points enough to 
determine the curve. 

Problem 72. — To draw the projections of the line of inter- 
section of two surfaces of revolution, axes not intersecting. 

Place the surfaces as before, one axis perpendicular 
to H and the other parallel to V. Cut the surfaces by 
horizontal planes. The elevations of the sections are 
straight lines, the plan of one a circle, of the other a curve 
which will generally have to be plotted by points, but only 
a small part of the curve need be drawn for each section. 



Exercises 

348: A is the axis of a cylindrical hole \y± in. in diameter. Show 
the projections of the line of intersection (Fig. 172). 

349. A pyramid with an altitude of 3 in. has a base 2 in. square, 
parallel to H with its diagonal parallel to V. 
It is intersected by a triangular hole of 1 in. 
side, the axis of which is 1 in. above the base, 
Y± in. from the axis of the pyramid and inclined 
at 30° to V. Show plan and elevation of the 
pyramid . 

350. A vertical tube of 2J^ in. outside di- 
ameter and X}/2 in. inside diameter has a hori- 
zontal cylindrical hole bored through it of 1% 
in. diameter. The distance between the axes is 
}£ in. Show the projection of the line of inter- 
section on a vertical plane to which the axis of the horizontal hole 
is inclined at 45°. 

351. Show the projections of the points in which the line A pierces 
the sphere (Fig. 173). 

352. A sphere of 3 in. diameter is intersected by a cylinder of 2 in. 
diameter- The axis of the cylinder is J^ in. from the center of the 



i 


5 


}if 


^0° 


_.,i 


i 


A" 


M 





Fig. 172. 



92 



DESCRIPTIVE GEOMETRY 



sphere. Show H and V projections of the line of intersection when 
the axis of the cylinder is parallel to H, inclined at 30° with V and in a 
horizontal plane with the center of the sphere. 

353. Show H, V and P projections of the line of intersection of the 
cylinders (Fig. 174). 

354. Show the projections of the lines of intersection of the solid 





Fig. 173. 



Fig. 174. 



of revolution of Exercise 344 with a sphere of 1% in. diameter, center 
of sphere on axis of solid and J^ in. from equatorial plane. 

355. Show the projections of the lines of intersection of a cylinder 
and^cone of revolution. The vertex angle of the cone is 60°, altitude 
23^ in. The diameter of the cylinder is 1 in., and the axis of the 
cylinder intersects the axis of the cone at a point l}i in. from the 
vertex, and at an angle of 75°. 

356. A surface generated by the revolution of a circular arc of 

120° about its chord, which is 3 in. long, 
is intersected by a cylinder of 1J^ in. di- 
ameter, the axes meeting at the middle 
point at an angle of 45°. Show the pro- 
jections of the lines of intersection. 

357. A connecting rod stub end of 
rectangular section is finished in a lathe 
to the outline indicated in Fig. 175. 
Show complete plan and elevation. 

358. An anchor ring generated by a 1 in. circle revolving about an 
axis % in. from its center is intersected by a cylinder of 1 in. diameter, 
the^axis of the cylinder meeting the axis of the surface at 30° and 




o 



g Fig. 175. ] 



INTERSECTION OF SURFACES 



93 



tangent to the inner surface of the ring. Show the projections of 
the line of intersection. 

359. Show the line of intersection of the cone and cylinder of 
Fig. 176. 





Fig. 176. 



Fig. 177. 



360. Show the line of intersection of the cone and cylinder of Fig. 
177. 

361. Show the projections of the line of intersection of the torus of 
Exercise 345 with a cylinder of 2 in. diameter, 

axis of cylinder parallel to axis of torus and 
1 in. from it. 

362. Two cylinders of revolution of 2 in. 
diameter intersect with the axis of each cyl- 
inder tangent to the other cylinder. One 
has its axis perpendicular to H, the other 
parallel to X. Show the projections of the 
curve of intersection. 

363. Points a, b and c form an equi- 
lateral triangle of 2% in. side lying in H. A 
point p is 1^ in. from a, 2 in. from b and 
2J^ in. from c. Show the projections of the 
point. 

364. Two spheres of 2% in. and 3 in. di- 
ameters have the plans of their centers X% 

in. apart on a line making 30° with X. The center of the smaller 
sphere is \\i in. above H, of the larger in H. Find the traces of the 
plane, the projections of the center and the radius of the circle in 
which the spheres intersect. 




Fig. 178. 



94 DESCRIPTIVE GEOMETRY 

365. A right cone of slant height 3 in., diameter of base 2J^ in., 
lies against H with the plan of the axis parallel to X. It is intersected 
by a cylinder of 1% in. diameter lying against H and inclined at 60° 
to V. The plan of the axis of the cylinder crosses the plan of the axis 
of the cone at a point 2 in. from its vertex. Show projections of the 
line of intersection. 

366. A sphere of 1)^ in. diameter has its center J^ in. above the 
center of a hemispherical shell of 4 in. diameter, as shown in Fig. 178. 
Find the shadow of the sphere on the concave surface of the shell. 

367. abc is a triangle in H. ab is 3 in., be is 4 in. and dc is 4J^ in 
d is a point in space. 6 for the line ad is 30°, for bd 45° and for cd 45° 
Show the projections of the point d. 

XX. Helicoidal Surfaces. 

A helix is the curve traced upon a cylinder of revolution 
by a point having uniform motion of rotation about the 
cylinder and at the same time a uniform motion of transla- 
tion along the elements of the cylinder. The curve therefore 
makes equal angles with the elements and when the surface 
of the cylinder is developed forms a straight line which is 
inclined to the developed elements at an angle whose tan- 
gent equals the circumference of the cylinder divided by 
the axial pitch of the helix. The axial pitch is the distance 

from one coil of the helix 
to the next, measured 
along an element, as in 
Fig. 179. 

The surface generated by 
a line moving in contact 
with a given helix and in 
some fixed relation to its axis, is called a helicoidal surface. 
Ordinary screw surfaces are helicoidal surfaces in which 
the generating line intersects the axis; if at right angles 
the surface is part of the square thread; if at 60°, the ordi- 
nary form of V thread. 

If a helicoidal surface is intersected by cylinders of 
revolution having the same axis, the lines of intersection 




HELIC01DAL SURFACES 



95 



are helices having the same axial pitch, but inclined to the 
base of the cylinder at angles whose tangents vary inversely 
as the diameter of the cylinder. 

In general helicoidal surfaces are not developable, and 
are sometimes called "skew screw surfaces." 

A single exception occurs in the case when a line moves 
so as to be in all positions tangent to the given helix. The 
surface thus generated can be developed, because eonsecu- 




Fig. ISO. 



tive elements are consecutive tangents to the helix, and 
therefore intersecting lines. This surface is called the 
developable helicoid. 

Figure 180 represents that portion of the surface included 
between two parallel planes at a distance apart equal to the 
pitch of the helix. If a card be cut in the form of the right 
triangle adc, having ad equal to the axial pitch of the helix 
and dc equal to the circumference of the cylinder of the helix ; 
and if this card be placed in the position indicated and rolled 



'■■ 


i 




^ 






V 


i 


/ 1 \r 


d 


^ ?;rr — 


,e , 



96 DESCRIPTIVE GEOMETRY 

about the cylinder in contact with it, the line ac will be in 
all positions in contact with the helix abd and will therefore 
generate the desired surface. Two like surfaces will be 
generated meeting at the helix in a cuspidal edge. The 
pdint a will trace on the upper plane an involute of the plan 
of the helix, while the point b will trace the opposite involute 

on the lower plane. Only 
the lower surface is shown 
in the figure. 

It can be shown that 
when the helicoidal surface 
is developed, the cuspidal 
edge will develop into a 
circular arc of radius R = r 
sec 2 6 , in which r is the ra- 
FlG 181 dius of the cylinder and 6 

the angle which an element 
makes with the plane of the base. The value of R can be 
found graphically by the triangle construction shown in 
Fig. 181. Lay off ee equal to r, draw ef perpendicular to 
ed and fg perpendicular to ca, then eg is equal to R. The 
line of intersection of the surface with a plane perpendic- 
ular to the axis will develop into an involute of the arc of 
radius R. 

Exercises 

368. A helicoidal surface is generated by a horizontal line in contact 
with a vertical axis and a helix of 2 in. pitch. Show the projections of 
the surface included between two horizontal planes 2 in. apart and 
two concentric cylinders of 1 in. and 2 in. diameters, axes coinciding 
with the axis of the helix. 

369. A screw is 2 in. outside diameter and has two square threads 
per inch. Show the projections of the portion included between two 
parallel planes 2 in. apart. 

370. A V thread of 1 in. pitch and 4 in. outside diameter is cut by a 
plane parallel to the axis and 1 in. from it. Show the section. 

371. Show the section of the V thread of the preceding exercise by 
a plane perpendicular to the axis. 



HYPERB0L01D OF REVOLUTION OF ONE NAPPE 97 



372. A cylinder 1J£ in. in diameter and 2 in. long has a helix of 2 in. 
pitch drawn on its surface. Show the projections of the surface 
generated by a line moving so as to be always tangent to the helix. 

373. A screw conveyor is of the form of a developable helicoid. 
The diameter of the inner helix is 12 in., of the outer one 30 in. and the 
pitch is 36 in. Show projections and development of one coil. Scale, 
1 in. = 1 ft, in. 

374. A pulley of 2 ft. in. diameter, 2 ft. 4 in. above a floor, 
is connected by a crossed belt 8 in. wide to a pulley of 2 ft. in. 
diameter and 1 ft. 2 in. below the floor. The pulleys are 4 ft. 
in. apart, center to center. Assuming that the belt in twisting takes 
the form of a helicoidal surface, show the lines of intersection with the 
floor. Scale, 1 in. =1 ft. in. 

XXI. Hyperboloid of Revolution of One Nappe. 

This surface may be generated (1) by the rotation of 
a straight line about an axis not in the same plane, (2) 
by the rotation of an hyperbola about its conjugate axis, 
(3) by a circle of variable radius, 
the center of which moves along 
the conjugate axis of an hyperbola, 
the plane of the circle being per- 
pendicular to this axis and the 
circumference of the circle always 
in contact with the hyperbola. The 
surface is best studied by con- 
sidering it as generated by the first 
method. 

Suppose the line be, Fig. 182, to 
rotate about the vertical axis 0, 
keeping at a constant distance Oa 
from it and making a constant angle 
6 with H. The line will take the 
successive positions indicated, the plan being tangent to 
the circle Oa, from which the elevations are determined. 

The surface thus generated will have circles for its cross 
sections, the smallest being the gorge circle of radius Oa. 




Fig. 182. 



98 DESCRIPTIVE GEOMETRY 

Any meridian section will be an hyperbola having as its 
conjugate axis. It will be seen that the same surface would 
have been generated if the line having the plan b h & had had 
the point c in the plane of the top and b in the plane of the 
lower base. Hence the surface is a doubly generated one, 

being made up of two sets of 
elements. It is the only sur- 
face of revolution which is 
doubly generated . Any 
point on the surface will lie 
on two straight line ele- 
ments which are, however, 
not of the same set. These 
two elements determine the 
plane tangent to the surface 
at this point. 

From the nature of the generation of this surface no two 
elements of the same set can intersect or be parallel, hence 
the surface is not developable. 




Fig. 183. 



30' 

mm 1 



Y/////// /M 



Exercises 

375. Show the projections of an hyperboloid of revolution having 
the radius of the gorge circle 1 in. and the elements inclined at 30° to 
the planes of the bases, which are 2 in. apart. Show also the lines of 
intersection by planes parallel to V and 
at distances of % in. and l}i in. from 
the axis. 

376. A point on the surface of the 
hyperboloid of the preceding exercise 
lies in a meridian plane at 60° to V and 
is 1J^ in. from the axis of the hyper- 
boloid. Show the traces of the plane 
tangent at this point. 

377. An hyperboloid is generated by revolving line B about A as 
axis (Fig. 183). Show the projections of another line which, when 
revolved about A, will generate the same surface. 

378. A rope passes through a doorway as shown in plan in Fig. 184. 




Fig. 184. 



PRACTICAL APPLICATIONS 



99 



The door is 3 ft. in. wide and the rope is inclined at 45° to the 
horizontal. Show where the door must be cub 'to allow it to be shut. 

379. A surface generated by rotating a cube of 2 in. edge about a 
diagonal, is cut by a plane parallel to the axis and 1 in. from it. Show 
the outline of the section. 

380. A square prism of 1 in. side is cut by a plane so that the 
section has angles of 60° and 120° and the long diagonal measures 
3 in. Show the section. 



XXII. Practical Applications. 

Gearing. — In the transmission of power by toothed gears 
the motion transmitted from one shaft to another is the 
same as that which would be transmitted by rolling of one 
surface on another. These surfaces, which are imaginary 
on an actual gear, are called the 
pitch surfaces. In the most com- 
mon case, that of spur gears, which 
are used when the two shafts are 
parallel, the pitch surfaces are cyl- 
inders. In special cases these 
cylinders may have right sections 
other than circular, as in the case 
of elliptical gearing. 

Two particular types of gears 
involve the principles of Descrip- 
tive Geometry in their design. 
Skew bevel gears used when the 
two shafts are neither parallel nor intersecting, have pitch 
surfaces which are hyperboloids of revolution. 

Suppose that A and B, Fig. 185, are the center lines of the 
two shafts that are to be connected by skew bevel gears. 
If some third line intermediate between A and B, such as 
Z), is revolved in turn about A and 5, it will generate two 
hyperboloids of revolution which will be in contact along 
the line D, and which will roll on each other as they are 




Fig. 185. 



100 



DESCRIPTIVE GEOMETRY 




rotated about the axes A and B. These hyperboloids 
therefore constitute the pitch surfaces of the required gears. 
The speed ratio of the two gears will depend on the posi- 
tion of the line D. The angles between the H projections 
D h and A h , and between D h and B h , may be called the skew 
angles of the shafts A and B. If the radii of the gorge 
circles of the hyperboloids are made in the ratio of the tan- 
gents of the skew angles, then the speed ratio will be the 
ratio of the sines of these angles. The problem will usually 
be in the following form: Given the two shafts and the 

required speed ratio, 
to determine the ele- 
ment of contact and 
the pitch surfaces. 

The following 
method by George B. 
Grant, M. E., is a 
direct solution. 1 

Suppose A and B, 
Fig. 186, to be the 
given shafts, and the 
required speed ratio 
m : n. Draw lines par- 
allel to A^and B h at distances from them whose ratio is m : n. 
The point of intersection of these lines will be a point on D h , 
as will also the point of intersection of A h and B h . The sum 
of the two gorge radii is the distance between A v and B v , 
and this is to be divided in the ratio of the tangents of the 
skew angles. Draw any line ab perpendicular to D h and 
through a draw a line ac at any convenient angle. Lay 
off ac equal to the distance from A v to B v , connect be and 
draw de parallel to be. ae and ec will then be the required 
gorge radii. D v is thus determined at a distance ae below 

1 For a complete discussion of this subject, see Mr. Grant's book 
"Gearing." 



S: "^ 




1 




v 


^ 


^ 


=21 


A r 


D" 


B v 



Fig. 186. 



PRACTICAL APPLICATIONS 



101 



A v . The hyperboloid pitch surfaces can then be drawn by 
the principles explained in the previous chapter. 

The other type of gearing whose design involves the prin- 
ciples of Descriptive Geometry is the elliptical gear which 
works with a gear of two or more lobes. The axes of the 
gears are parallel, so that the pitch surfaces are cylinders 
(not cylinders of revolution). Since all right sections of 
cylinders are equal, this problem may be simplified by con- 
sidering a section, in which case the pitch surfaces become 
pitch lines. 

It can be shown that an ellipse formed by a plane section 
of a cone of revolution may be rolled on the development 
of the cone, the axis of rotation for the ellipse being taken 
through one focus, and for the cone development through 
the developed vertex. The ellipse and the cone develop- 
ment will therefore serve as pitch lines for the required 
gears. In order to have practi- 
cable gears, the pitch lines must 
form closed curves. That is, the 
development of the cone must 
occupy some aliquot part of 360°. 
In other words the gear which 
mates with the ellipse may consist 
of two lobes of 180° each, or three 
lobes of 120° each, etc. This 
means, according to the equation 
given in the chapter on develop- 
ments, that the ratio of the slant height of the cone to the 
radius of the base must be a whole number and equal to 
the number of lobes on the gear. 

The following method, due to Prof. John B. Peddle, 
determines the position of the secant plane which will 
cut an ellipse of required major and minor axis from a 
cone of revolution. 

Draw the triangle abc, Fig. 187, representing the cone 




102 



DESCRIPTIVE GEOMETRY 



with vertex at a and base be, the length be being the major 
axis of the ellipse. The ratio of the slant height ae to the 
radius oe is made equal to the desired number of lobes. 
Circumscribe a circle about the triangle. Locate the foci 
of the ellipse, as / and /'. Produce ao to d, and draw a 
circular arc with d as center and db as radius. With center 
at b and radius //' strike an arc to 
e. Join be and produce to g. The 
distances gb and ge will then be the 
longest and shortest elements of the 
cone. 

The problem of properly shaping 
and locating the teeth on the pitch 
surfaces of a gear has no connection 



Fig. 188. 




Fig. 189. 



with Descriptive Geometry and will not be considered. 

Belt Drive. — In the transmission of power by belts, it is 
necessary that the center line of the belt approaching a 
pulley must lie in the center plane of the pulley. In Fig. 188 
the center plane of the pulley is represented by the line ab, 
and the center line of the belt approaching the pulley must 
lie along this line. If it were along some other line, as 
ed, the belt would run off the pulley. On the other side of 



PRACTICAL APPLICATIONS 103 

the pulley, where the belt leaves, it does not have to obey 
this law. 

It is sometimes necessary to connect two shafts which 
are not parallel. This can be easily accomplished for one 
direction of rotation by properly locating the pulleys. One 
such case is illustrated in Fig. 189. It will be seen by trac- 
ing out the path of the belt that the center line of the belt 
approaching either pulley lies in the center plane of that 
pulley. If the direction of motion is reversed this is no 
longer true, and the belt will run off the pulleys. In order 
to make the drive reversible it is necessary to provide one 
or two guide pulleys, which will so guide the belt that it 
will run in either direction. The location of a single guide 
pulley to accomplish this is a problem in Descriptive 
Geometry. 

As an illustration of this, assume that the shafts A and 
B in Fig. 190 are to be connected by a belt so as to run 
in the directions indicated, or with both directions reversed. 
The pulleys may be so placed that one side of the belt 
lies in the middle planes of both pulleys, and consequently 
will run in either direction. The other side of the belt will 
require a guide pulley, and this must be so placed that 
it will guide the portions of the belt on each side of it in 
the planes of the pulleys on A and B. The center lines of 
these two portions of the belt will be lines lying in the mid- 
dle planes of the pulleys and intersecting in a point which 
is common to both the planes. 

The first step in the solution of the problem, after the 
main pulleys have been located, is to select the position of 
this point lying in the planes of both pulleys. It may be 
taken at any convenient height, but not too close to either 
A or B. In Fig. 190, plane 3 is the mid plane of pulley 
S, plane 4- that of pulley A, and the point is chosen at c, 
on the line of intersection of 3 and 4- From this point 
the two lines C and D are drawn, one tangent to each 



104 



DESCRIPTIVE GEOMETRY 




Fig. 190. 



PRACTICAL APPLICATIONS 105 

pulley. This may or may not require the revolution of 
the mid planes of the pulleys into V or H, according as the 
pulleys are or are not oblique to V and H. In the figure, 
plane 3 has been revolved into V and plane 4 into H in 
order to draw lines C and D. The plane of the lines C and 
D is the plane of the guide pulley. Find the traces of this 
plane, as 5 h and 5 V . Revolve into H or V, and locate the 
revolved positions of point c and lines C and D. These are 
shown as (c), (C) and (D). 

Assuming that the size of the guide pulley is known, it 
can be drawn tangent to (C) and (D) and its center point 
(d) located in revolved position. Counter revolve plane 5 
and obtain d h and d v . The axis of the guide pulley shaft 
may then be drawn as line E, through d and perpendicular 
to plane 5. 

In order to find the projections of the guide pulley aux- 
iliary planes may be used to advantage. To construct the 
V projection take plane 6 containing line E and perpendicu- 
lar to V. Revolve it into V and construct the revolved 
position of the section cut from the pulley, which will be a 
rectangle. Counter revolve plane 6, which gives the major 
and minor axes of the ellipses forming the required V 
projection. The H and P projections may be found by a 
similar construction. 

Roof Truss Problems. — In the design of roof trusses a num- 
ber of angles and dimensions may be found by the methods 
of Descriptive Geometry. 1 In steel trusses standard shapes 
such as I beams, angles, channels, etc., are used, and it is 
chiefly in the design of connecting angle plates that the 
methods of Descriptive Geometry offer convenient graphical 
solutions. 

1 Messrs. H. L. McKibben and L. E. Gray, in a book entitled "Hip 
and Valley Design" give graphical methods as well as analytic 
formula for the solution of a number of types of roof, hopper and 
pipe line designs. 



106 



DESCRIPTIVE GEOMETRY 



Figure 191 shows the plan of a hip roof on a rectangular 
building. The lines marked A are the hip rafters and those 
marked B are the purlins. The purlins are so placed that 

their flanges lie in the plane of 
the roof and their webs are per- 
pendicular to this plane. The 
hip rafters have their webs in 
a vertical plane. If the four 
planes of the roof are equally 
inclined to the horizontal, the hip 
rafter plans will be 45° lines. 
As an illustration of a prob- 
lem relating to this type of roof, let it be required to de- 
sign a bent plate for connecting a purlin to a hip rafter, 
the plate to be riveted to the web of each. 




Fig. 191. 




Fig. 192. 

In Fig. 192, A is the hip rafter and B the purlin. The 



PRACTICAL APPLICATIONS 107 

plane of the hip rafter web is vertical, and its traces are 3 h 
and 3 V . The plane of the purlin web is perpendicular to 
the roof plane and its traces 2 h and -2 V may be found by 
means of a P projection, 2 V being perpendicular to A p . 
The line C, which is the line of intersection of these two 
planes will be the line along which the connecting plate 
must be bent, and the angle between the planes is the angle 
to which the plate must be bent. 

The shape of the connecting plate can be determined by 
revolving planes 2 and 3 into or parallel to V. The dimen- 
sions of the plate will be determined by the size and shape 
of the structural shapes used for hip rafter and purlin. 
Assuming these to be known we may proceed as follows: 
Revolve plane 3, and with it lines A and C, into V. The 
lines in the revolved positions are shown as (A) and (C). 
Revolve plane 2 parallel to V about B as axis. Line B 
remains stationary and fine C revolves to tV- To deter- 
mine the shape of the plate before bending draw any line 
representing B. Draw line C making with B the angle 
between B v and d v . Draw A making with C the angle be- 
tween (A) and (C). A and B then represent the developed 
positions of these lines, and C the bend line. 

It must be remembered that A and B lie in the plane 
of the roof, while the connecting plate must be at a dis- 
tance from the roof plane so as to make allowance for the 
flanges of the rafter and purlin. These allowances, marked 
x and y in the figure, will be determined by the shape and 
dimensions of the structural shapes used. Draw parallels 
to A and B at these distances from them, and then lay off 
the required widths m and n of the two portions of the plate. 
These lines show the limiting lines for the required plate, 
and any outline lying inside of these, and bent along the 
line C will serve the purpose, provided sufficient plate area 
is provided for the required rivet holes. The angle to 
which the plate must be bent is the angle between planes 2 



108 DESCRIPTIVE GEOMETRY 

and 3, and may be determined by the methods of Prob- 
lem 46. The projections of the bent plate can then be 
drawn. 

Moulding Cutter Knives. — Wooden mouldings are cut by 
means of rotating knives, and the shape of the moulding 
cut depends upon the shape of the cutting edges of the 
knives. In general the shape of the moulding is not the 
same as that of the knife, because the axis of rotation does 
not lie in the plane of the knife. The knife, in revolving, 
sweeps out a solid of revolution, the meridian section of 
which is the shape of the moulding. If the plane of the 
knife contained the axis, then the shape of the knife and 
moulding would be the same. 

If we know the shape of the moulding desired, the dis- 
tance of the knife plane from the axis of rotation and also 
the distance of some particular point of the cutting edge 
from the axis, w r e can find the shape of the knife as follows: 
Draw a solid of revolution whose meridian section is the re- 
quired moulding shape, making the particular point at the 
required distance from the axis. Cut this solid by a plane 
parallel to the axis and at a distance from it equal to the 
distance from the knife plane to the axis. This section is 
the required knife shape. 

Mining Problems. — Deposits of ore frequently occur in 
beds which are essentially plane. The direction of a hori- 
zontal line in the ore bed is called the strike, and the angle 
of inclination of the bed with the horizontal is called the 
dip. The line along which the ore bed intersects the surface 
of the ground is called the outcrop. The locations and 
elevations of three points on the outcrop are sufficient 
data to determine the plane of the ore bed, and the various 
problems connected with this subject can be easily solved 
graphically. 

To Find the Strike. — First locate both H and V projections 
of the three given points. Through one of them draw a 



PRACTICAL APPLICATIONS 109 

line parallel to H and lying in the plane of the three points. 
The H projection of this line gives the strike. This line 
can be most conveniently located without finding the traces 
of the plane. The V projection is drawn parallel to X. 
Join the other two points by a line, find both projections 
of the point where the horizontal line crosses it and join 
the H projection of this point with the H projection of the 
first point, which gives the H projection required. 

To Find the Dip. — Take a P plane perpendicular to the 
horizontal line just found. Find the P projections of the 
three given points, and join them by a straight line. This 
represents the P projection or trace of the plane of the ore 
bed, and its inclination gives the required angle. 

To Find the Depth of a Shaft. — Suppose the location and 
elevation of a fourth point on the surface of the ground 
is given, to find the depth of a shaft to strike the ore bed. 
Locate the P projection of this point. Its distance verti- 
cally over the ore bed is shown in the P projection. 

For other problems and methods of this sort, the student 
is referred to the article "The Application of Descriptive 
Geometry to Mining Problems/' by Joseph W. Roe, 
Transactions American Institute of Mining Engineers, 
Vol. XLI, page 512. 

Exercises 

381. Two horizontal shafts lie in planes 2 in. apart and their plans 
are inclined at 60°. It is desired to connect these shafts by skew bevel 
gearing whose speed ratio is 1 : 2. Find the radii of the gorge circles 
of the hyperboloids which will be the pitch surfaces of the gears, also 
the projections of the element of contact. 

382. Show the projections of the hyperboloids of the preceding 
exercise. 

383. A gear has a pitch line of elliptic form, major axis 3 in. and 
minor axis 2 in. Determine the pitch line of a bilobe gear which will 
work with the elliptic gear. 

384. Determine the pitch line of a trilobe gear which will work 
with the elliptic gear of the preceding exercise. 



110 



DESCRIPTIVE GEOMETRY 



B 



~^A V 



385. Two horizontal shafts, whose center lines are shown in Fig. 
193, are to be connected by a belt, and are to run in the directions 
indicated or both reversed. The pulley on shaft A is to be 18 in. in 
diameter, that on B 14 in., and the guide pulley 15 in. The guide 
pulley is to be located about halfway between the shafts. The width 
of belt is to be 3 in. Determine the location of the guide pulley and 

show projections of pulleys and 
belt. Scale, 1 in. = 1 ft. in. 
This exercise requires a drawing 
space about 12 in. X 18 in. to in- 
clude a P projection. 

386. A rectangular building has 
a hip roof whose hip rafters are 
inclined at 30° to the horizontal. 
It is required to design a bent 
plate connection between purlin 
and hip rafter, given the following 
data : Connecting plate to be riveted 
to the webs of rafter and purlin. 
The rafter is a 6 in. I beam which 
will allow a plate 4 in. wide, and the 
purlin is a 4 in. channel, and 
since the plate is to be riveted to 
its back side, no allowance need 
be made for its flanges. Two 
%-in. rivets are to be used in each 
portion of the plate , and center 
distances of rivets are to be at least 
2 in. No rivet center is to come 
closer than 1 in. to the edge of the plate nor closer than 1J^ in. to 
the bend line. The rivets on the channel must be at least 1J^ m - 
from edge of channel, to allow for flanges. Draw the developed 
plate, locate bend line and determine bending angle. Scale, 3 in. = 1 
ft. in. 

387. Find the profile for a cutter knife which shall cut a moulding 
of the section shown in Fig. 194. Assume that the plane of the knife 
is \Y2 in. from the axis of rotation, and that the nearest point of the 
cutting edge is 2 in. from the axis. 

388. Take the same data as in the preceding exercise, but the 
moulding section shown in Fig. 195. 

389. Given three points a, b and c, on the outcrop of a bed of ore 
(Fig. 196). The elevations of the points are as follows: 



^ 
> 



B v 



Fig. 193. 



IMPOSSIBLE AND INDETERMINATE EXERCISES 111 

a, 2920 ft. 

b, 3060 ft, 

c, 2840 ft. 

Find the strike and dip. Scale. 1 in. = 80 ft. 

390. Taking the same data as in the preceding exercise, find the 





Fig. 194. 



Fig. 195. 




Fig. 196. 

depth of a shaft at d to strike the ore bed. Point d is 250 ft. due east 
of a, and its elevation is 2890 ft. 

391. Solve Exercise 390 by finding the traces of the plane abc and 
finding the point of intersection of the vertical shaft with this plane. 



XXIII. Impossible and Indeterminate Exercises. 

In the following an attempt should be made to solve the 
exercise. Then state clearly why the exercise is impossible 
or indeterminate. 

392. Find the traces of the plane containing lines C and D (Fig. 
75. page 46). 

393. Find the traces of a plane containing line A and perpendicular 
to line B (Fig. 125. page 63). 

394. Find a line of plane 2 making 60° with V (Fig. 118, page 60). 



112 



DESCRIPTIVE GEOMETRY 



395. Find the traces of a plane which makes 30° with H and 45° 
with V. 

396. Find the traces of the plane containing points a, b and c 
(Fig. 197). 

397. Pass a plane through line B and parallel to line A (Fig. 119, 
page 60). 







» — 


k 


rM 








r 




^4 


sf 


*^ ' 








/ > 












( f 


N 


' i 




- 


f V 


\ f 


K— /* ' 


k i 


/"-> 


^J 


L. 




^ / 


> ! < 


^ 


J 


V. | 


c ijfc 


L 








Fig 


. 197. 









398. Find the projections of a line making 45° with H and 60° 
with V. 

399. Through line A pass a plane making 30° with H (Fig. 119, 
page 60). 

400. Show the traces of the plane containing point a and perpen- 
dicular to plane 2 (Fig. 118, page 60). 



LIBRARY OF CONGRESS 




003 572 690 1 



